Idea Transcript
P4 Stress and Strain
Dr. A.B. Zavatsky HT08
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.
1
Stress Transformation Equations y1
σy
σy1
τyx
τy1x1
τxy
y x
σx
y
θ σx1
τxy
τx1y1
τyx
σ x1 = τ x1 y1 =
2
+
σ x −σ y 2
( σ x −σ y ) − sin 2θ 2
x1 σx1
σx
σ x +σ y
τx1y1
x τy1x1 σy1
cos 2θ + τ xy sin 2θ
+ τ xy cos 2θ
If we vary θ from 0° to 360°, we will get all possible values of σx1 and τx1y1 for a given stress state. It would be useful to represent σx1 and τx1y1 as functions of θ in graphical form. 2
To do this, we must re-write the transformation equations.
σ x1 −
σ x +σ y 2
=
τ x1 y1 =
σ x −σ y 2
cos 2θ + τ xy sin 2θ
( σ x −σ y ) − sin 2θ
+ τ xy cos 2θ
2
Eliminate θ by squaring both sides of each equation and adding the two equations together.
⎛ ⎜ σ x1 − ⎜ ⎝
σ x +σ y ⎞ 2
2
⎛σ x −σ y ⎞
⎟ + τ x1 y12 = ⎜ ⎟ ⎜ ⎠ ⎝
⎟ ⎟ ⎠
2
2
+ τ xy 2
Define σavg and R
σ avg =
σ x +σ y 2
R=
⎛σ x −σ y ⎞ ⎜ ⎜ ⎝
2
⎟ ⎟ ⎠
2
+ τ xy 2 3
Substitue for σavg and R to get
(σ x1 −
σ avg ) 2 + τ x1 y12 = R 2
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 (1835-1918). 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. 4
Sign Convention for Mohr’s Circle y1 σy1
(σ x1 −
y τy1x1
τx1y1
x1 σx1
θ σx1 τx1y1
σ avg ) 2 + τ x1 y12 = R 2 2θ σavg
x
σx1 R
τy1x1 σy1
τx1y1
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. 5
Procedure for Constructing Mohr’s Circle 1. 2. 3.
4.
5.
6.
Draw a set of coordinate axes with σx1 as abscissa (positive to the right) and τx1y1 as ordinate (positive downward). Locate the centre of the circle c at the point having coordinates σx1 = σavg and τx1y1 = 0. 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°. 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°. 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). Using point c as the centre, draw Mohr’s circle through points A and B. This circle has radius R. (based on Gere) 6
B
σy
τyx τxy
y
σy
x
σx
B (θ=90)
τxy
-τxy
σx
A τyx
σx1
c R
τxy
A (θ=0) σavg τx1y1
σx 7
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)
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σy
B σy1
x
σx
σx
τxy
2θ+180
E (θ+90)
A τyx
-τx1y1
σx1
c D (θ) R
2θ
y1 σy1
A (θ=0)
y τy1x1
τx1y1
x1 σx1
E
D
θ
σx1 σx1
τx1y1
τxy
y
B (θ=90)
τx1y1
τyx
τx1y1
x
τy1x1 σy1 9
Principal Stresses
σy
B
τyx τxy
y
B (θ=90)
R
A τyx
σx1
σ1
c
σx
τxy
2θp2
σ2
x
σx
2θp1
y σ2
A (θ=0)
P2
θp2
σ1
P1
θp1
x
σ1
τx1y1
σ2 10
Maximum Shear Stress
B
σy
τyx τxy
y
B (θ=90)
x
σx
τmin
σx
τxy τyx
2θs σx1
c τmax
A
R
Note carefully the directions of the y shear forces.
σs
A (θ=0) σs τx1y1
τmax τmax
σs τmax
θs
σs
τmax σs
11
x
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. c = σ avg = R=
σ x +σ y 2
=
− 80 + 50 = −15 2
σ 1,2 = c ± R σ 1,2 = −15 ± 69.6
(50 − (− 15)) 2 + (25)2
R = 65 2 + 25 2 = 69.6
σ 1 = 54.6 MPa σ 2 = −84.6 MPa
A (θ=0)
σ2
σ1 c
B
50 MPa
R
σ
B (θ=90)
y 80 MPa
80 MPa
x
A
τ max = R = 69.6 MPa σ s = c = −15 MPa
τmax
25 MPa 50 MPa
τ
12
50 MPa
25 = 0.3846 80 − 15 2θ 2 = 21.0° tan 2θ 2 =
y 80 MPa
80 MPa
x
2θ1 = 21.0 + 180° = 201°
θ1 = 100.5° θ 2 = 10.5°
25 MPa 50 MPa
A (θ=0)
σ2
σ1
y
c
2θ2
R
54.6 MPa
2θ1 o
100.5 84.6 MPa
σ
B (θ=90)
84.6 MPa
10.5o
54.6 MPa
x
2θ τ
13
50 MPa
2θ 2 = 21.0° 2θ s min = − (90 − 21.0) = −69.0°
y 80 MPa
θ s min = −34.5°
80 MPa
x
τmin
25 MPa
taking sign convention into account
50 MPa
A (θ=0)
R
15 MPa 15 MPa
σ
c
2θ2
y
2θ
2θsmin
2θsmax 55.5
B (θ=90)
o
-34.5o
x 15 MPa
τmax
2θ 2 = 21.0° 2θ s max = 21.0 + 90° = 111.0°
15 MPa 69.6 MPa
τ
θ s max = 55.5° 14
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° clockwise and draw the corresponding stress elements. 50 MPa
C (θ = -30°)
y 80 MPa
80 MPa
x
-60°
A (θ=0) 25 MPa
σx1 = c – R cos(2θ2+60) σy1 = c + R cos(2θ2+60) τx1y1= -R sin (2θ2+60) σx1 = -26 σy1 = -4 τx1y1= -69
50 MPa
y1
y 25.8 MPa
σ
2θ2
4.15 MPa
B (θ=90)
D -60+180° x
o
-30
25.8 MPa 4.15 MPa
C 68.8 MPa
2θ
D (θ = -30+90°)
x1
τ
θ = -30° 2θ = -60° 15
Principal Stresses σ1 = 54.6 MPa, σ2 = -84.6 MPa But 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 MPa σ2 = 0 MPa σ3 = -84.6 MPa This means three Mohr’s circles can be drawn, each based on two principal stresses:
A (θ=0)
σ3
σ2
σ1
σ
B (θ=90)
σ1 and σ3 σ1 and σ2 σ2 and σ3
τ
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σ3
σ1
σ1
σ3
σ1
σ3
σ3
σ1
σ2
σ1
σ
σ3
σ3 σ1
σ1
σ3
τ 17
B
The stress element shown is in plane stress. What is the maximum shear stress?
σy
τxy
y x
σx
σx
τxy
B
σ3
τyx
A τyx
σ1
σ2
σx1 σ −σ2 τ max(1,2) = 1
A
τx1y1
2 σ −σ3 σ 2 τ max(2,3) = 2 = 2 2
σ1 − σ 3 σ1 τ = = overall maximum max(1,3) 2
2
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Introduction to the Stress Tensor y
σyy
τyx
τyz
τxy
τzy
σxx
τzx τxz σzz z
σyy
σxx x
⎛ σ xx τ xy τ xz ⎞ ⎜ ⎟ ⎜ τ yx σ yy τ yz ⎟ ⎜⎜ ⎟⎟ τ τ σ zx zy zz ⎠ ⎝
Normal stresses on the diagonal Shear 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. 19
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,
⎛ σ xx τ xy τ xz ⎞ ⎜ ⎟ ⎜ τ yx σ yy τ yz ⎟ ⎜⎜ ⎟⎟ ⎝ τ zx τ zy σ zz ⎠
0 ⎞ ⎛ σ1 0 ⎜ ⎟ ⎜ 0 σ2 0 ⎟ ⎜0 ⎟ 0 σ 3⎠ ⎝
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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Example: The state of plane stress at a point is represented by the stress element below. Find the principal stresses. 50 MPa
y 80 MPa
80 MPa
x
25 MPa 50 MPa
⎛ σ x τ xy ⎞ ⎛ − 80 − 25 ⎞ ⎟ = ⎜⎜ ⎟⎟ M = ⎜⎜ ⎟ ⎝τ yx σ y ⎠ ⎝ − 25 50 ⎠ 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. 21
⎛ − 80 − λ − 25 ⎞ ⎟⎟ = 0 det ⎜⎜ 50 − λ ⎠ ⎝ − 25 (−80 − λ )(50 − λ ) − (−25)(−25) = 0
λ2 + 30λ − 4625 = 0 λ = −84.6, 54.6
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.
⎛ − 80 − λ − 25 ⎞ ⎛ x ⎞ ⎛ 0 ⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ 50 − λ ⎠ ⎝ y ⎠ ⎝ 0 ⎠ ⎝ − 25 − 25 ⎞ ⎛ x ⎞ ⎛ 0 ⎞ ⎛ − 80 − 54.6 ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ 50 − 54.6 ⎠ ⎝ y ⎠ ⎝ 0 ⎠ ⎝ − 25
⎛ − 134.6 − 25 ⎞ ⎛ x ⎞ ⎛ 0 ⎞ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎜⎜ − 4.64 ⎠ ⎝ y ⎠ ⎝ 0 ⎠ ⎝ − 25 x = −0.186 y ⎛ − 0.186 ⎞ ⎟⎟ is one eigenvector. ⎜⎜ ⎝ 1 ⎠ 22
⎛ − 80 − λ − 25 ⎞ ⎛ x ⎞ ⎛ 0 ⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ 50 − λ ⎠ ⎝ y ⎠ ⎝ 0 ⎠ ⎝ − 25 − 25 ⎛ − 80 − (−84.6) ⎞ ⎛ x ⎞ ⎛0⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ − 25 50 − (−84.6) ⎠ ⎝ y ⎠ ⎝ 0 ⎠ ⎝
⎛ 4.6 − 25 ⎞ ⎛ x ⎞ ⎛ 0 ⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ − 25 134.6 ⎠ ⎝ y ⎠ ⎝ 0 ⎠ x = 5.388 y ⎛ 5.388 ⎞ is the other eigenvector. ⎜⎜ ⎟⎟ ⎝ 1 ⎠
Before finding the angles at which the principal stresses act, we can check to see if the eigenvectors are correct. 0 ⎞ ⎛ 54.6 ⎟⎟ D = ⎜⎜ − 84.6 ⎠ ⎝ 0
⎛ − 0.186 5.388 ⎞ ⎟⎟ C = ⎜⎜ 1 ⎠ ⎝ 1
⎛ − 80 − 25 ⎞ ⎟⎟ M = ⎜⎜ ⎝ − 25 50 ⎠
D = C −1M C
C −1 =
1 where A = matrix of co - factors AT det C ⎛ − 0.179 0.967 ⎞ − 1 ⎟⎟ C = ⎜⎜ ⎝ 0.179 0.033 ⎠ 23
0 ⎞ ⎛ − 0.179 0.967 ⎞ ⎛ − 80 − 25 ⎞ ⎛ − 0.186 5.388 ⎞ ⎛ 54.6 ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ D = ⎜⎜ 1 ⎠ ⎝ 0 − 84.6 ⎠ ⎝ 0.179 0.033 ⎠ ⎝ − 25 50 ⎠ ⎝ 1 To find the angles, we must calculate the unit eigenvectors:
⎛ − 0.186 ⎞ ⎛ − 0.183 ⎞ ⎟⎟ ⎟⎟ → ⎜⎜ ⎜⎜ ⎝ 1 ⎠ ⎝ 0.983 ⎠
⎛ 5.388 ⎞ ⎛ 0.938 ⎞ ⎟⎟ ⎟⎟ → ⎜⎜ ⎜⎜ ⎝ 1 ⎠ ⎝ 0.183 ⎠
And then assemble them into a rotation matrix R so that det R = +1.
⎛ 0.983 − 0.183 ⎞ ⎟⎟ R = ⎜⎜ ⎝ 0.183 0.983 ⎠
det R = (0.983)(0.983) − (0.183)(−0.183) = 1
The rotation matrix has the form
⎛ cosθ R = ⎜⎜ ⎝ sin θ
− sin θ ⎞ ⎟⎟ cosθ ⎠
D ′ = RT M R
So θ = 10.5°, as we found earlier for one of the principal angles. 24
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).
D ′ = RT M R ⎛ 0.983 0.183 ⎞ ⎛ − 80 − 25 ⎞ ⎛ 0.983 − 0.183 ⎞ ⎟⎟ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ D′ = ⎜⎜ ⎝ − 0.183 0.983 ⎠ ⎝ − 25 50 ⎠ ⎝ 0.183 0.983 ⎠ 0 ⎞ ⎛ − 84.6 ⎟⎟ D′ = ⎜⎜ 54.6 ⎠ ⎝ 0 y 54.6 MPa o
100.5 84.6 MPa
84.6 MPa
10.5o
x
So, the transformation equations, Mohr’s circle, and eigenvectors all give the same result for the principal stress element.
54.6 MPa
25
Finally, we can use the rotation matrix approach to find the stresses on an inclined element with θ = -30°.
⎛ cos(−30°) − sin( −30°) ⎞ ⎛ 0.866 0.5 ⎞ ⎟⎟ = ⎜⎜ ⎟⎟ R = ⎜⎜ ⎝ sin( −30°) cos(−30°) ⎠ ⎝ 0.5 0.866 ⎠ M ′ = RT M R ⎛ 0.866 − 0.5 ⎞ ⎛ − 80 − 25 ⎞ ⎛ 0.866 0.5 ⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ M ′ = ⎜⎜ ⎝ 0.5 0.866 ⎠ ⎝ − 25 50 ⎠ ⎝ − 0.5 0.866 ⎠ ⎛ − 25.8 − 68.8 ⎞ ⎛ σ x1 τ xy ⎞ ⎟ ⎟⎟ = ⎜⎜ M ′ = ⎜⎜ ⎟ ⎝ − 68.8 − 4.15 ⎠ ⎝ τ yx σ y1 ⎠
25.8 MPa
Again, the transformation equations, Mohr’s circle, and the stress tensor approach all give the same result.
y1
y
4.15 MPa
o
-30
x 25.8 MPa x1
4.15 MPa 68.8 MPa
26