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
Mohr’s Circle for Plane Stress Transformation equations for plane stress. Procedure for constructing Mohr’s circle. Stresses on an inclined element. Principal stresses and maximum shear stresses. Introduction to the stress tensor.
2
Stress Transformation Equations
σx σx
σy
τxy
τyx
τxy
τyx
xy
x
y
σx1
θ
y1
x1
σx1
σy1
σy1
τx y1 1 τy x1 1
τx y1 1
τy x1 1
( )θτθ
σστ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+−
−=
+−
++
=
If we vary θ from 0° to 360°, we will get all possible values of σx1 and τx1y1for a given stress state. It would be useful to represent σx1 and τx1y1 as functions of θ in graphical form.
3
To do this, we must rewrite the transformation equations.
( )θτθ
σστ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+−
−=
+−
=+
−
Eliminate θ by squaring both sides of each equation and adding the two equations together.
22
211
2
1 22 xyyx
yxyx
x τσσ
τσσ
σ +⎟⎟⎠
⎞⎜⎜⎝
⎛ −=+⎟⎟
⎠
⎞⎜⎜⎝
⎛ +−
Define σavg and R
22
22 xyyxyx
avg R τσσσσ
σ +⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
+=
4
Substitue for σavg and R to get
( ) 2211
21 Ryxavgx =+− τσσ
which is the equation for a circle with centre (σavg,0) and radius R.
This circle is usually referred to as Mohr’s circle, after the German civil engineer Otto Mohr (18351918). He developed the graphical technique for drawing the circle in 1882.
The construction of Mohr’s circle is one of the few graphical techniques still used in engineering. It provides a simple and clear picture of an otherwise complicated analysis.
5
Sign Convention for Mohr’s Circle
x
y
σx1
θ
y1
x1
σx1
σy1
σy1
τx y1 1 τy x1 1
τx y1 1
τy x1 1
( ) 2211
21 Ryxavgx =+− τσσ
σx1
τx1y1
R
2θσavg
Notice that shear stress is plotted as positive downward.
The reason for doing this is that 2θ is then positive counterclockwise, which agrees with the direction of 2θ used in the derivation of the tranformation equations and the direction of θ on the stress element.
Notice that although 2θ appears in Mohr’s circle, θ appears on the stress element.
6
Procedure for Constructing Mohr’s Circle
1. Draw a set of coordinate axes with σx1 as abscissa (positive to the right) and τx1y1 as ordinate (positive downward).
2. Locate the centre of the circle c at the point having coordinates σx1 = σavg and τx1y1 = 0.
3. Locate point A, representing the stress conditions on the x face of the element by plotting its coordinates σx1 = σx and τx1y1 = τxy. Note that point A on the circle corresponds to θ = 0°.
4. Locate point B, representing the stress conditions on the y face of the element by plotting its coordinates σx1 = σy and τx1y1 = −τxy. Note that point B on the circle corresponds to θ = 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 Mohr’s circle through points Aand B. This circle has radius R.
(based on Gere)
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σx σx
σy
τxy
τyx
τxy
τyx
xy
σx1
τx1y1
σavg
cR
A (θ=0)
σx
τxy
AB (θ=90)
σy
τxy
B
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Stresses on an Inclined Element 1. On Mohr’s circle, measure an angle 2θ counterclockwise from
radius cA, because point A corresponds to θ = 0 and hence is the reference point from which angles are measured.
2. The angle 2θ locates the point D on the circle, which has coordinates σx1 and τx1y1. 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 2θ + 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 θ, the point on Mohr’s circle corresponding to the x1 face moves counterclockwise through an angle 2θ.
(based on Gere)
9
σx σx
σy
τxy
τyx
τxy
τyx
xy
A (θ=0)
σx1
τx1y1
c
R
B (θ=90)A
B
x
y
σx1
θ
y1
x1
σx1
σy1
σy1
τx y1 1 τy x1 1
τx y1 1
τy x1 1
2θ
D
D (θ)
σx1
τx1y1
σy1
τx1y1
2θ+180E (θ+90)
E
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Principal Stresses
σx σx
σy
τxy
τyx
τxy
τyx
xy
A (θ=0)
σx1
τx1y1
c
R
B (θ=90)A
B
x
y
σ1
θp1
σ1
σ2
σ2
θp2
P1
2θp1
P2
2θp2
σ1σ2
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Maximum Shear Stress
σx σx
σy
τxy
τyx
τxy
τyx
xy
A (θ=0)
σx1
τx1y1
c
R
B (θ=90)
A
B
2θs
σs
τmax
τmin
x
y
σs
θs
σs
σs
σs
τmaxτmax
τmax
τmax
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 Mohr’s 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
σ
τ
152
50802
−=+−
=+
== yxavgc
σσσ
c
A (θ=0)
A
B (θ=90)B
( )( ) ( )
6.692565
251550
22
22
=+=
+−−=
R
R
R
σ1σ2
MPa6.84MPa6.54
6.6915
2
1
2,1
2,1
−==
±−=
±=
σσ
σ
σ Rc
τmaxMPa15
MPa6.69
s
max−====
cR
στ
13
84.6 MPa
84.6 MPa
54.6 MPa
x
y
54.6 MPa
10.5o
100.5o
σ
τ
c
A (θ=0)
B (θ=90)R
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
σ1σ2
2θ2
2θ1
°=°=
°=°+=°=
=−
=
5.105.100
2011800.2120.212
3846.01580
252tan
21
1
2
2
θθ
θθ
θ
2θ
14
σ
τ
c
A (θ=0)
B (θ=90)R
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
2θ2
2θ
15 MPa
15 MPa
15 MPa
x
y
15 MPa
69.6 MPa
34.5o
55.5o
2θsmax
°=
°=°+=°=
5.550.111900.212
0.212
max
max
2
s
sθ
θθ
2θsmin
taking sign convention into account
°−=
°−=−−=°=
5.340.69)0.2190(2
0.212
min
min
2
s
sθ
θθ
τmax
τmin
15
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPaA (θ=0)
σ
τ
B (θ=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 (θ = 30°)
CD (θ = 30+90°)
D
2θ2
σx1 = c – R cos(2θ2+60)σy1 = c + R cos(2θ2+60)
τx1y1= R sin (2θ2+60)σx1 = 26
σy1 = 4τx1y1= 69
2θ
θ = 30°2θ = 60°
16
σ
τ
A (θ=0)
B (θ=90)
Principal Stresses σ1 = 54.6 MPa, σ2 = 84.6 MPaBut we have forgotten about the third principal stress!
Since the element is in plane stress (σz = 0), the third principal stress is zero.
σ1 = 54.6 MPaσ2 = 0 MPaσ3 = 84.6 MPa
σ1σ2σ3This means three Mohr’s circles can be drawn, each based on two principal stresses:
σ1 and σ3
σ1 and σ2
σ2 and σ3
17
σ1
σ3
σ1
σ3
σ
τ
σ1σ2σ3
σ1
σ3
σ1
σ3
σ1σ1σ3
σ3
18
σx σx
σy
τxy
τyx
τxy
τyx
xy
The stress element shown is in plane stress.What is the maximum shear stress?
A
B
σx1
τx1y1
A
B
σ3 σ1σ2
22131
)3,1max(σσστ =
−=
221
)2,1max(σστ −
=
22232
)3,2max(σσστ =
−=
overall maximum
19
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
zzzyzx
yzyyyx
xzxyxx
στττστττσ
y
z
xσxx σxx
σyy
σyy
σzz
τxy
τyx
τxzτzx
τzy
τyz
Introduction to the Stress Tensor
Normal stresses on the diagonalShear stresses off diagaonal
τxy = τyx, τxz = τzx, τyz = τzy
The normal and shear stresses on a stress element in 3D can be assembled into a 3x3 matrix known as the stress tensor.
20
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
στττστττσ
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
32
1
000000
σσ
σ
In mathematical terms, this is the process of matrix diagonalization in which the eigenvalues of the original matrix are just the principal stresses.
21
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
σττσ
We must find the eigenvalues of this matrix.
Remember the general idea of eigenvalues. We are looking for values of λ such that:Ar = λr where r is a vector, and A is a matrix.Ar – λr = 0 or (A – λI) r = 0 where I is the identity matrix.
For this equation to be true, either r = 0 or det (A – λI) = 0.Solving the latter equation (the “characteristic equation”) gives us the eigenvalues λ1 and λ2.
22
6.54,6.840462530
0)25)(25()50)(80(
05025
2580det
2
−==−+
=−−−−−−
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−
−−−
λλλ
λλλ
λ
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 – λI ) r= 0 one at a time and solve for r.
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−−−−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−−
−−−
00
6.545025256.5480
00
50252580
yx
yx
λλ
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−−−−
1186.0
186.000
64.425256.134
yxyx
is one eigenvector.
23
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−−−
−−−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−−
−−−
00
)6.84(502525)6.84(80
00
50252580
yx
yx
λλ
⎟⎟⎠
⎞⎜⎜⎝
⎛
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
−
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
24
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞⎜⎜⎝
⎛−
−−⎟⎟⎠
⎞⎜⎜⎝
⎛−=
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=′⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
θθθθ
cossinsincos
The rotation matrix has the form
So θ = 10.5°, as we found earlier for one of the principal angles.
25
⎟⎟⎠
⎞⎜⎜⎝
⎛−=′
⎟⎟⎠
⎞⎜⎜⎝
⎛ −⎟⎟⎠
⎞⎜⎜⎝
⎛−
−−⎟⎟⎠
⎞⎜⎜⎝
⎛−
=′
=′
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, Mohr’s circle, and eigenvectors all give the same result for the principal stress element.
26
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−−−
=′
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−
−−⎟⎟⎠
⎞⎜⎜⎝
⎛ −=′
=′
11
15.48.688.688.25
866.05.05.0866.0
50252580
866.05.05.0866.0
yyxxyx
T
M
M
RMRM
σττσ
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛°−°−°−−°−
=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 θ = 30°.
Again, the transformation equations, Mohr’s circle, and the stress tensor approach all give the same result.