Lecture 6 Mohr's Circle for Plane Stress

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Transcript of Lecture 6 Mohr's Circle for Plane Stress

1
P4 Stress and Strain Dr. A.B. ZavatskyHT08
Lecture 6
Mohrs Circle for Plane Stress Transformation equations for plane stress. Procedure for constructing Mohrs circle. Stresses on an inclined element. Principal stresses and maximum shear stresses. Introduction to the stress tensor.

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Stress Transformation Equations
x x
y
xy
yx
xyyx
xy
x
y
x1
y1
x1
x1
y1
y1
x y1 1 y x1 1
x y1 1 y x1 1
( )
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+
=
+
++
=
If we vary from 0 to 360, we will get all possible values of x1 and x1y1for a given stress state. It would be useful to represent x1 and x1y1 as functions of in graphical form.

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To do this, we must rewrite the transformation equations.
( )
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+
=
+
=+
Eliminate by squaring both sides of each equation and adding the two equations together.
22
211
2
1 22 xyyx
yxyx
x
+
=+
+
Define avg and R
22
22 xyyxyx
avg R
+
=
+=

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Substitue for avg and R to get
( ) 221121 Ryxavgx =+ which is the equation for a circle with centre (avg,0) and radius R.
This circle is usually referred to as Mohrs circle, after the German civil engineer Otto Mohr (18351918). He developed the graphical technique for drawing the circle in 1882.
The construction of Mohrs circle is one of the few graphical techniques still used in engineering. It provides a simple and clear picture of an otherwise complicated analysis.

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Sign Convention for Mohrs Circle
x
y
x1
y1
x1
x1
y1
y1
x y1 1 y x1 1
x y1 1 y x1 1
( ) 221121 Ryxavgx =+
x1
x1y1
R
2avg
Notice that shear stress is plotted as positive downward.
The reason for doing this is that 2 is then positive counterclockwise, which agrees with the direction of 2 used in the derivation of the tranformation equations and the direction of on the stress element.
Notice that although 2 appears in Mohrs circle, appears on the stress element.

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Procedure for Constructing Mohrs Circle
1. Draw a set of coordinate axes with x1 as abscissa (positive to the right) and x1y1 as ordinate (positive downward).
2. Locate the centre of the circle c at the point having coordinates x1 = avg and x1y1 = 0.
3. Locate point A, representing the stress conditions on the x face of the element by plotting its coordinates x1 = x and x1y1 = xy. Note that point A on the circle corresponds to = 0.
4. Locate point B, representing the stress conditions on the y face of the element by plotting its coordinates x1 = y and x1y1 = xy. Note that point B on the circle corresponds to = 90.
5. Draw a line from point A to point B, a diameter of the circle passing through point c. Points A and B (representing stresses on planes at 90 to each other) are at opposite ends of the diameter (and therefore 180 apart on the circle).
6. Using point c as the centre, draw Mohrs circle through points Aand B. This circle has radius R.
(based on Gere)

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x x
y
xy
yx
xyyx
xy
x1
x1y1
avg
cR
A (=0)
x
xy
AB (=90)
y
xy
B

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Stresses on an Inclined Element 1. On Mohrs circle, measure an angle 2 counterclockwise from
radius cA, because point A corresponds to = 0 and hence is the reference point from which angles are measured.
2. The angle 2 locates the point D on the circle, which has coordinates x1 and x1y1. Point D represents the stresses on the x1 face of the inclined element.
3. Point E, which is diametrically opposite point D on the circle, is located at an angle 2 + 180 from cA (and 180 from cD). Thus point E gives the stress on the y1 face of the inclined element.
4. So, as we rotate the x1y1 axes counterclockwise by an angle , the point on Mohrs circle corresponding to the x1 face moves counterclockwise through an angle 2.
(based on Gere)

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x x
y
xy
yx
xyyx
xy
A (=0)
x1
x1y1
c
R
B (=90)A
B
x
y
x1
y1
x1
x1
y1
y1
x y1 1 y x1 1
x y1 1 y x1 1
2
D
D ()
x1
x1y1
y1
x1y1
2+180E (+90)
E

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Principal Stresses
x x
y
xy
yx
xyyx
xy
A (=0)
x1
x1y1
c
R
B (=90)A
B
x
y
1 p1
1
2
2
p2P1
2p1
P2
2p2
12

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Maximum Shear Stress
x x
y
xy
yx
xyyx
xy
A (=0)
x1
x1y1
c
R
B (=90)
A
B
2s
s
max
min
x
y
s s
s
s
s
maxmax
max
max
Note carefully the directions of the
shear forces.

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Example: The state of plane stress at a point is represented by the stress element below. Draw the Mohrs circle, determine the principal stresses and the maximum shear stresses, and draw the corresponding stress elements.
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
152
50802
=+
=+
== yxavgc
c
A (=0)
A
B (=90)B
( )( ) ( )
6.692565
251550
22
22
=+=
+=
R
R
R
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MPa6.84MPa6.54
6.6915
2
1
2,1
2,1
==
=
=
Rc
maxMPa15
MPa6.69
s
max====
cR

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84.6 MPa
84.6 MPa
54.6 MPa
x
y
54.6 MPa
10.5o
100.5o
c
A (=0)
B (=90)R
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
12
22
21
==
=+==
=
=
5.105.100
2011800.2120.212
3846.01580
252tan
21
1
2
2
2

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c
A (=0)
B (=90)R
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
22
2
15 MPa
15 MPa
15 MPa
x
y
15 MPa
69.6 MPa
34.5o55.5o
2smax
=
=+==
5.550.111900.212
0.212
max
max
2
s
s
2smin
taking sign convention into account
=
===
5.340.69)0.2190(2
0.212
min
min
2
s
s
max
min

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80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPaA (=0)
B (=90)25.8 MPa
25.8 MPa
4.15 MPa
x
y
4.15 MPa68.8 MPa
x1
y1
30o
Example: The state of plane stress at a point is represented by the stress element below. Find the stresses on an element inclined at 30 clockwiseand draw the corresponding stress elements.
60
60+180
C ( = 30)
CD ( = 30+90)
D
22
x1 = c R cos(22+60)y1 = c + R cos(22+60)
x1y1= R sin (22+60)x1 = 26
y1 = 4x1y1= 69
2
= 302 = 60

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A (=0)
B (=90)
Principal Stresses 1 = 54.6 MPa, 2 = 84.6 MPaBut we have forgotten about the third principal stress!
Since the element is in plane stress (z = 0), the third principal stress is zero.
1 = 54.6 MPa2 = 0 MPa3 = 84.6 MPa
123This means three Mohrs circles can be drawn, each based on two principal stresses:
1 and 3
1 and 2
2 and 3

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1
3
1
3
123
1
3
1
3
113
3

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x x
y
xy
yx
xyyx
xy
The stress element shown is in plane stress.What is the maximum shear stress?
A
B
x1
x1y1
A
B
3 12
22131
)3,1max( ==
221
)2,1max( =
22232
)3,2max( ==
overall maximum

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zzzyzx
yzyyyx
xzxyxx
y
z
xxx xx
yy
yy
zz
xy
yx
xzzx
zy
yz
Introduction to the Stress Tensor
Normal stresses on the diagonalShear stresses off diagaonal
xy = yx, xz = zx, yz = zy
The normal and shear stresses on a stress element in 3D can be assembled into a 3x3 matrix known as the stress tensor.

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From our analyses so far, we know that for a given stress system, it is possible to find a set of three principal stresses. We also know that if the principal stresses are acting, the shear stresses must be zero. In terms of the stress tensor,
zzzyzx
yzyyyx
xzxyxx
32
1
000000
In mathematical terms, this is the process of matrix diagonalization in which the eigenvalues of the original matrix are just the principal stresses.

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80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
Example: The state of plane stress at a point is represented by the stress element below. Find the principal stresses.
=
=
50252580
yyxxyxM
We must find the eigenvalues of this matrix.
Remember the general idea of eigenvalues. We are looking for values of such that:Ar = r where r is a vector, and A is a matrix.Ar r = 0 or (A I) r = 0 where I is the identity matrix.
For this equation to be true, either r = 0 or det (A I) = 0.Solving the latter equation (the characteristic equation) gives us the eigenvalues 1 and 2.

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6.54,6.840462530
0)25)(25()50)(80(
05025
2580det
2
==+
=
=
So, the principal stresses are 84.6 MPa and 54.6 MPa, as before.
Knowing the eigenvalues, we can find the eigenvectors. These can be used to find the angles at which the principal stresses act. To find the eigenvectors, we substitute the eigenvalues into the equation (A I ) r= 0 one at a time and solve for r.
=
=
00
6.545025256.5480
00
50252580
yx
yx
=
=
1186.0
186.000
64.425256.134
yxyx
is one eigenvector.

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=
=
00
)6.84(502525)6.84(80
00
50252580
yx
yx
=
=
1388.5
388.500
6.13425256.4
yxyx
is the other eigenvector.
Before finding the angles at which the principal stresses act, we can check to see if the eigenvectors are correct.
CMCD
MCD
1
50252580
11388.5186.0
6.84006.54
=
=
=
=
=
==
033.0179.0967.0179.0
factorsco ofmatrix wheredet
1
1
1
C
AAC
C T

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=
=
6.84006.54
11388.5186.0
50252580
033.0179.0967.0179.0
D
To find the angles, we must calculate the unit eigenvectors:
183.0938.0
1388.5
983.0183.0
1186.0
And then assemble them into a rotation matrix R so that det R = +1.
1)183.0)(183.0()983.0)(983.0(det983.0183.0183.0983.0
==
= RR
RMRDR T=
=
cossinsincos
The rotation matrix has the form
So = 10.5, as we found earlier for one of the principal angles.

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=
=
=
6.54006.84
983.0183.0183.0983.0
50252580
983.0183.0183.0983.0
D
D
RMRD T
Using the rotation angle of 10.5, the matrix M (representing the original stress state of the element) can be transformed to matrix D (representing the principal stress state).
84.6 MPa
84.6 MPa
54.6 MPa
x
y
54.6 MPa
10.5o
100.5o
So, the transformation equations, Mohrs circle, and eigenvectors all give the same result for the principal stress element.

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=
=
=
=
11
15.48.688.688.25
866.05.05.0866.0
50252580
866.05.05.0866.0
yyxxyx
T
M
M
RMRM
=
=866.05.0
5.0866.0)30cos()30sin()30sin()30cos(
R
25.8 MPa
25.8 MPa
4.15 MPa
x
y
4.15 MPa68.8 MPa
x1
y1
30o
Finally, we can use the rotation matrix approach to find the stresses on an inclined element with = 30.
Again, the transformation equations, Mohrs circle, and the stress tensor approach all give the same result.