failure criteria: mohr's circle and principal
TRANSCRIPT
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• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
Third EditionLECTURE
227.4
Chapter
FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES
byDr. Ibrahim A. Assakkaf
SPRING 2003ENES 220 – Mechanics of Materials
Department of Civil and Environmental EngineeringUniversity of Maryland, College Park
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 1ENES 220 ©Assakkaf
Example 2The stresses shown in Figure 12a act at a point on the free surface of a stressed body. Determine the normal stresses σnand σt and the shearing stress τnt at this point if they act on the rotated stress element shown in Figure 12b.
The Stress Transformation Equations for Plane Stress
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 2ENES 220 ©Assakkaf
The Stress Transformation Equations for Plane Stress
Example 2 (cont’d)
MPa 70
MPa 40
MPa 10
nσtσ ntτ
015
ntFigure 12
(a) (b)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 3ENES 220 ©Assakkaf
The Stress Transformation Equations for Plane Stress
Example 2 (cont’d)The given values are as follows:
000t
0 1059015 ,15
MPa 40 MPa, 70 MPa, 10
=+==
+=−=−=
θθ
τσσ
n
xyyx
nσtσ ntτ
015
nt
t
015
090n
3
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 4ENES 220 ©Assakkaf
The Stress Transformation Equations for Plane Stress
Example 2 (cont’d)Applying Eq. 12 for the given values
( ) ( ) ( ) ( )
(Tension) MPa 981.5MPa 981.5
15cos15sin)40(215sin7015cos10
cossin2sincos22
22
==
+−−=
++=
n
xyyxn
σ
θθτθσθσσ
( ) ( ) ( ) ( )
n)(comprssio MPa 86MPa 98.85
105cos105sin)40(2105sin70105cos10
cossin2sincos22
22
=−=
+−−=
++=
t
xyyxt
σ
θθτθσθσσ
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 5ENES 220 ©Assakkaf
The Stress Transformation Equations for Plane Stress
Example 2 (cont’d)( ) ( )( ) ( ) ( )( )
MPa 64.1915sin15cos40)15cos()15sin()70(10
sincoscossin22
22
=−+−−−−=
−+−−=
nt
xyyxnt
τ
θθτθθσστ
MPa 70
MPa 40
MPa 10
MPa 98.5=nσ
MPa 86=tσ MPa 64.19=ntτ
015
nt
4
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 6ENES 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 τnton 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).
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 7ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Principal Stresses
x
y
z
(b)
ksi 7
ksi 12
ksi 25⋅ o
(a)
⋅
Surfaces perpendicular to z-axisare stress-free.
Figure 13
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 8ENES 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 stress-free 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.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 9ENES 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.
6
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 10ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Variation of Stresses as Functions of θ
-20
-10
0
10
20
30
40
0 60 120 180 240 300 360
Angle θ (Degrees)
Stre
ss (K
si)
Stress snStress tnt
nt
n
τσ
ksi 12
ksi 7
ksi 25x
y nt
θ
Figure 14
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 11ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Principal Stresses– Principal Stresses for Special Loading
Conditions:• Bar under axial load
• Shaft under Pure BendingA
PAP
2 and maxmax == τσ (14)
JcTmax
maxmax ==τσ (15)
7
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 12ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Principal Stresses for Axially Loaded Bar
θ
P
PF
Original Area, AInclined Area, An
Figure 15
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 13ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Principal Stresses for Axially Loaded Bar
PN
Vθ
θ
θ
θθ
cos
sincos
AA
PVPN
n =
==
θ
( )θθ
θ
θσ 2cos12
cos
cos
cos 2 +====A
PAP
AP
AN
nn
Figure 16a
8
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 14ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Principal Stresses for Axially Loaded Bar
PN
Vθ
θ
θ
θθ
cos
sincos
AA
PVPN
n =
==
θ
θθθ
θ
θτ 2sin2
cossin
cos
sinA
PAP
AP
AV
nn ====
Figure 16b
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 15ENES 220 ©Assakkaf
Principal Stresses for Axially Loaded Bar
σn is maximum when θ = 00 or 1800
τn is maximum when θ = 450 or 1350
– Also
Therefore2max
maxστ =
AP
AP
2 and maxmax == τσ
(16)
(17)
Principal Stresses and Maximum Shearing Stress
9
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 16ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Principal Stresses for Shaft under Pure Torsion
x
y A
x
y
xyτxyτ
yxτ
yxτ
(a)
(b)x
yt
α
α
n
σn dA
τn t dA
α
τyx dA sin α
τ x y
dA
cosα
(c)
Fig.17
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 17ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Principal Stresses for Shaft under Pure Torsion yt
α
n
σn dA
τn t dA
α
τyx dA sin α
τ x y
dA
cosα
( ) ατααττ 2cossincos 22xyxynt =−=
(18)αταατσ 2sincossin2 xyxyn ==
(19)
JcTmax
maxmax ==τσ (20)
10
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 18ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Development of Principal Stresses Equations
Recall Eq 13
θτθσσ
τ 2cos2sin2 xy
yxnt +
−−=
θτθσσσσ
σ 2sin2cos22 xy
yxyxn +
−+
+= (13a)
(13b)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 19ENES 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
( ) 0 2cos22sin
2sin2cos22
=+−−=
+
−+
+=
θτθσσ
θτθσσσσ
θθ
xyyx
xyyxyxn
dd
ddσ
2
tan2or 2
2tan
or
1
−=
−= −
yx
xyp
yx
xyp σσ
τθ
σστ
θ (21)
set
11
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 20ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Development of Principal Stresses Equations
Substituting the expression for 2θp into Eq. 13a, yields
Eq. 21 gives the two principal stresses in the xy-plane, and the third stress σp3 = σz = 0.
22
2 ,1 22 xyyxyx
pp τσσσσ
σ +
−±
+= (22)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 21ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Principal StressesPrincipal stresses σmax and σmin can be computed from
where subscript p refers to the planes of maximum and minimum values of σn.
22
2 ,1 22 xyyxyx
pp τσσσσ
σ +
−±
+= (22a)
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 22ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Location of the Plane of Principal Stresses
2
tan21 1
−= −
yx
xyp σσ
τθ (22b)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 23ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Notes on Principal Stresses Equation1. 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.
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 24ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Notes on Principal Stresses Equation3. 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. 25ENES 220 ©Assakkaf
Development of Maximum Shearing Stress Equation
Recall Eq 13b: θτθσσ
τ 2cos2sin2 xy
yxnt +
−−=
( ) 0 2sin22cos
2cos2sin2
=−−−=
+
−−=
θτθσσ
θτθσσ
θθτ
xyyx
xyyxnt
dd
dd
2
tan2or 2
2tan
or
1
−=
−−= −
xy
yx
xy
yx
τσσ
θτσσ
θ ττ
set
(23)
Principal Stresses and Maximum Shearing Stress
14
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 26ENES 220 ©Assakkaf
Development of Principal Shearing Stress Equation
Substituting the expression for 2θτ into Eq. 13b, yields
Eq. 24 gives the maximum in-plane shearing stress.
22
2 xyyx
p τσσ
τ +
−±= (24)
Principal Stresses and Maximum Shearing Stress
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 27ENES 220 ©Assakkaf
Maximum In-Plane Shearing StressMaximum in-plane shearing stress can be computed from
where the subscript p refers to the plane of maximum in-plane shearing stress τp.
(24a)22
2 xyyx
p τσσ
τ +
−±=
Principal Stresses and Maximum Shearing Stress
15
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 28ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Location of the Plane of Maximum Shearing Stress
2
tan21 1
−= −
xy
yx
τσσ
θτ (24b)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 29ENES 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.
16
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 30ENES 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. 31ENES 220 ©Assakkaf
pθ
xyτ
xσ
pθ
yσ
1pσ
2pσ
nσ
maxτ
maxτ
nσ
1pσ
2pσ
21 Pp σσ >
045
045
x
y Wedge-shaped BlockFig.18
Principal Stresses and Maximum Shearing Stress
17
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 32ENES 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
• 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.
221 pp
p
σστ
−=
yxpp σσσσ +=+ 21
(25)
(26)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 33ENES 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:( ) ( ) ( ) 2/0max ,2/0max ,2/max 2121 pppp σσσσ −=−=−=
18
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 34ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Example 3Normal 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. 35ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Example 3 (cont’d)
ksi 4
ksi 12
ksi 6
Fig.19The given values for usein Eqs. 22 and 24 are:
σx = +12 ksiσy = - 4 ksiτxy = - 6 ksi
19
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 36ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Example 3 (cont’d)Using Eq. 22a for the given values:
Therefore,
( ) ( ) 10462
4122
)4(12
22
22
22
±=−+
−−
±−+
=
+
−±
+= xy
yxyxp τ
σσσσσ
0
(C) ksi 6 ksi 6104(T) ksi 14ksi 14104
3
2
1
==
=−=−==+=+=
zp
p
P
σσ
σσ
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 37ENES 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
The location θp of the principal stresses is computed from Eq. 22b
( ) ksi 102
202
6142
21max +==
−−=
−= pp σσ
τ
( )( )
01 18.43 412
62tan21
=
−−−
= −
yxpθ
20
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 38ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Example 3 (cont’d)Sketch of principal and max shearing stressesksi 6
ksi 12
043.18
ksi 4
ksi 10
ksi 4
ksi 14
ksi 6
045
x
y
ksi 6
maxksi 10 ττ ==p
(T) ksi 42
4122
=−
=+
= yxn
σσσ
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 39ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Example 4Normal 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
21
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 40ENES 220 ©Assakkaf
Principal Stresses and Maximum Shearing Stress
Example 4 (cont’d)
MPa 36
MPa 72
MPa 24
Fig.20The given values for usein Eqs. 22 and 24 are:
σx = +72 MPaσy = +36 MPaτxy = - 24 MPa
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 41ENES 220 ©Assakkaf
Example 4 (cont’d)Using Eq. 22a for the given values:
Therefore,
( ) ( ) 3054242
36722
)36(72
22
22
22
±=−+
+−
±++
=
+
−±
+= xy
yxyxp τ
σσσσσ
0
(T) MPa 24 ksi 243054(T) MPa 84ksi 843054
3
2
1
==
=+=−==+=+=
zp
p
P
σσ
σσ
Principal Stresses and Maximum Shearing Stress
22
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 42ENES 220 ©Assakkaf
Example 4 (cont’d)Since the σp1 and σp2 have the same sign, the maximum shearing stress is
The location θp of the principal stresses is computed from Eq. 22b
MPa 422
842
0842
01max +==
−=
−= pσ
τ
011 57.263672
)24(2tan212
tan21
−=
−−
=
−= −−
yx
xyp σσ
τθ
Principal Stresses and Maximum Shearing Stress
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 43ENES 220 ©Assakkaf
Example 4 (cont’d)Sketch of principal and max shearing stressesksi 6
MPa 72
057.26
MPa 36
MPa 30
MPa 54
MPa 84
MPa 24
045
x
y
MPa 24
MPa 42
MPa 302
24842
max
21
max
=
=−
=−
=
≠
τ
σστ
ττ
ppp
p
(T) MPa 542
36722
=+
=+
= yxn
σσσ
045
MPa 42
MPa 84
03 =pσ
MPa 42
Principal Stresses and Maximum Shearing Stress
23
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 44ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
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) Slide No. 45ENES 220 ©AssakkafMaximum Shearing Stress
Maximum shearing stress occurs for avex σσ =′
2
45by fromoffset
and 90by separated angles twodefines :Note
22tan
2
o
o
22
max
yxave
p
xy
yxs
xyyxR
σσσσ
θ
τσσ
θ
τσσ
τ
+==′
−−=
+
−==
24
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 46ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
IntroductionPoint, 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.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 47ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Plane Stresses using Mohr’s Circle• Recall Eqs. 13a and 13b,
• Squaring both equations, adding, and simplifying gives
θτθσσ
τ 2cos2sin2 xy
yxnt +
−−=
θτθσσσσ
σ 2sin2cos22 xy
yxyxn +
−+
+= (13a)
(13a)
22
22
22 xyyxyx
n ntτ
σστ
σσσ +
−=+
+− (27)
25
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 48ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
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 τaxis, and the radius of the circle is given by
22
2 xyyxR τ
σσ+
−= (28)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 49ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
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
26
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 50ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Plane Stresses using Mohr’s Circleplotted 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) Slide No. 51ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
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.
x
y
n
t
θ
yσ
xσ
xyτ
yxτ
xyτ
yxτxσ
yσ
A
A
θ
27
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 52ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Mohr’s Circle
yxτ
yσ
2pσ
2yx σσ +
xσnσ
1pσ
τ
σθ2
pθ2
pτ
ntτ
xyτ2yx σσ −
),( yxyH τσ
),( xyxV τσ
Fig.21
C
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 53ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Mohr’s Circle
Maximum shearing stress occurs for avex σσ =′
2
45by fromoffset and 90by separated angles twodefines :Note
22tan
2
o
o
22
max
yxave
p
xy
yxs
xyyxR
σσσσ
θ
τσσ
θ
τσσ
τ
+==′
−−=
+
−==
28
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 54ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Plane Stresses using Mohr’s CircleDrawing 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) Slide No. 55ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Plane Stresses using Mohr’s CircleDrawing 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) Slide No. 56ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
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) Slide No. 57ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Sign Conventions
σ
τ
(a) Clockwise → Above
σ
τ
(b) Counterclockwise → Below
Fig.22
The following jingle maybe helpful in rememberingthis conventions:“In the kitchen, the clockis above, and the counteris below.”
Beer and Johnston (1992)
σ
τ•
στ
•
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 58ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Points of Interests on Mohr’s Circle1. Point D that provides the principal stress
σp1.2. Point E that gives the principal stress σp2.3. Point A that provides the maximum in-
plane shearing stress -τp and the accompanied normal stress σavg that acts on the plane.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 59ENES 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 stressplot the points X and Y and construct the circle centered at C.
xyyx τσσ ,,
22
22 xyyxyx
ave R τσσσσ
σ +
−=
+=
• The principal stresses are obtained at A and B.
yx
xyp
ave R
σστ
θ
σσ
−=
±=
22tan
minmax,
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. 60ENES 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. 61ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress• Mohr’s circle for centric axial loading:
0, === xyyx AP τσσ
AP
xyyx 2=== τσσ
• Mohr’s circle for torsional loading:
JTc
xyyx === τσσ 0 0=== xyyx JTc τσσ
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 62ENES 220 ©Assakkaf
Mohr’s Circle for Plane StressExample 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.
ksi 14
ksi 20B
A
125Fig.23
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 63ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Example 5 (cont’d)
The given values for use indrawing Mohr’s circle are:
( ) ksi 89.14 )7.0(31724.45 cos
32
1420radius
ksi 172
1420
76.134512tan22 0
ksi 14
ksi 20
013
2
1
=−=−=
=−
==
=+
=
−=
−===
==
==
−
RC
R
C
n
pz
py
px
σ
θσσ
σσ
σσ
076.13424.45
3=R
σ
τ C
3=Rnσ
( ) ( ) ksi 13.2 24.45sin324.45sin === Rntτ
ntτ
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 64ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
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) Slide No. 65ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
Example 6 (cont’d)
SOLUTION:
• Construction of Mohr’s circle( ) ( )
( ) ( ) MPa504030
MPa40MPa302050
MPa202
10502
22 =+==
==−=
=−+
=+
=
CXR
FXCF
yxave
σσσ
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 66ENES 220 ©Assakkaf
Example 6 (cont’d)• Principal planes and stresses
5020max +=+== CAOCOAσ
MPa70max =σ
5020max −=−== BCOCOBσ
MPa30max −=σ
°=
==
1.53230402tan
p
p CPFX
θ
θ
°= 6.26pθ
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 67ENES 220 ©Assakkaf
Example 6 (cont’d)
• Maximum shear stress
°+= 45ps θθ
°= 6.71sθ
R=maxτ
MPa 50max =τ
aveσσ =′
MPa20=′σ