Mohr's Circle for Plane Stress

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P4 Stress and Strain
Dr. A.B. Zavatsky HT08
Lecture 6 Mohrs Circle for Plane StressTransformation 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.
1
Stress Transformation Equationsyy
y1
y y1x1 x1y1 x1 x1
yx xyx
y1
x xy yx
xx1 x1y1
x y1x1 y1
x1 = x1 y1 =
x + y2
+
x y2
( x y ) sin 2 2
cos 2 + xy sin 2
+ xy cos 2
If we vary from 0 to 360, we will get all possible values of x1 and x1y1 for a given stress state. It would be useful to represent x1 and x1y1 as functions of in graphical form.2
To do this, we must rewrite the transformation equations.
x1
x + y2
=
x y2
x1 y1 =
( x y ) sin 2 2
cos 2 + xy sin 2 + xy cos 2
Eliminate by squaring both sides of each equation and adding the two equations together.
x1
x + y 2
2
+ x1 y12 =
x y 2
2
+ xy 2
Define avg and R
avg =
x + y2
R=
x y 2
2
+ xy 23
Substitue for avg and R to get
( x1
avg ) 2 + x1 y12 = R 2
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.4
Sign Convention for Mohrs Circley1 y1 y y1x1 x1y1 x1 x1 x1y1 y1x1 y1 x x1
( x1
avg ) 2 + x1 y12 = R 22 avgR
x1
x1y1
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.5
Procedure for Constructing Mohrs Circle1. 2. 3. Draw a set of coordinate axes with x1 as abscissa (positive to the right) and x1y1 as ordinate (positive downward). Locate the centre of the circle c at the point having coordinates x1 = avg and x1y1 = 0. 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. 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. 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 Mohrs circle through points A and B. This circle has radius R. (based on Gere)6
4.
5.
6.
B y B (=90) xy c xy R
yy
yx xyx
x xy yx
x
A
x1
A (=0) avg x1y1 x7
Stresses on an Inclined Element1. 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. 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. 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. 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)
2.
3.
4.
8
B y1 B (=90) E (+90) x1y1 x1y1 c D () R 2 A (=0) x1x1 y1 y1x xy yx
yy x
yx xy x
2+180
A
x1y y1x1 x1y1 x1 x1
E
Dy1x1 y19
x
x1y1
x1y1
Principal StressesB (=90) 2p2x xy
B
yy x
yx xy x
Ayx
2
c R
1 2p1 A (=0)
x1y 2
P21
p2
1p1
P1
x
x1y1
210
Maximum Shear StressB (=90) min 2s c max Rsx xy
B
yy x
yx xy x
Ayx
x1
Note carefully the directions of the y shear forces.
A (=0) s x1y1
max max s ss
s max max11
x
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.c = avg =R=
x + y2
=
80 + 50 = 15 2
1,2 = c R 1,2 = 15 69.6 1 = 54.6 MPa 2 = 84.6 MPa
(50 ( 15)) 2 + (25)2A (=0)
R = 65 2 + 25 2 = 69.6
2 c B80 MPa 50 MPa y x 80 MPa
1 RB (=90)
A25 MPa 50 MPa
max
max = R = 69.6 MPa s = c = 15 MPa12
50 MPa
y 80 MPa x 80 MPa
25 = 0.3846 80 15 2 2 = 21.0 tan 2 2 = 21 = 21.0 + 180 = 201
25 MPa 50 MPa
1 = 100.5 2 = 10.5
A (=0)
2y54.6 MPa
122 21
c RB (=90)
100.5 84.6 MPa
o
84.6 MPa
10.5o
x
2
54.6 MPa
13
50 MPa
2 2 = 21.0 2 s min = (90 21.0) = 69.0
y 80 MPa x 80 MPa
s min = 34.5
25 MPa 50 MPa
min
taking sign convention into account
A (=0)
2smin
2 RB (=90)
y15 MPa 15 MPa
22 2smax55.5o
c
34.5o
x15 MPa
max
2 2 = 21.0 2 s max = 21.0 + 90 = 111.0
15 MPa 69.6 MPa
s max = 55.514
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
y 80 MPa x 80 MPa
C ( = 30) 60
A (=0)25 MPa 50 MPa
x1 = c R cos(22+60) y1 = c + R cos(22+60) x1y1= R sin (22+60) x1 = 26 y1 = 4 x1y1= 69
y 25.8 MPa
y1 4.15 MPa
22 D 60+180
B (=90)
30
o
x 25.8 MPa
D ( = 30+90)
2 = 30 2 = 6015
4.15 MPa
C68.8 MPa
x1
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 MPa 2 = 0 MPa 3 = 84.6 MPa This means three Mohrs circles can be drawn, each based on two principal stresses: 1 and 3 1 and 2 2 and 3
A (=0)
3
2
1
B (=90)
16
3
1
1
3
3
3
2
1
1 3 1
1
3
1
3
17
The stress element shown is in plane stress. What is the maximum shear stress?x
B
yy x
yx xy x
B
xy yx
A
3
2
1
x1 2 max(1,2) = 1
A
2 3 2 max(2,3) = 2 = 2 2
x1y1
max(1,3) = 1 3 = 1 overall maximum2 218
Introduction to the Stress Tensory
yy yz xx zy
yx xy
zx xz zzz
xx x
xx xy xz yx yy yz zx zy zz
yy
Normal stresses on the diagonal Shear 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.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,
xx xy xz yx yy yz zx zy zz
0 1 0 0 2 0 0 0 3
In mathematical terms, this is the process of matrix diagonalization in which the eigenvalues of the original matrix are just the principal stresses.
20
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 x 80 MPa
x xy 80 25 = M = yx y 25 50 We must find the eigenvalues of this matrix.
25 MPa 50 MPa
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 2.21
80 25 =0 det 25 50 (80 )(50 ) (25)(25) = 0
2 + 30 4625 = 0 = 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 I ) r = 0 one at a time and solve for r.
80 25 x 0 = 25 50 y 0 25 x 0 80 54.6 = 25 50 54.6 y 0
134.6 25 x 0 = 25 4.64 y 0 x = 0.186 y 0.186 1 is one eigenvector. 22
80 25 x 0 = 25 50 y 0 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= 0 84.6 D = C 1M C 0.186 5.388 C = 1 1 80 25 M = 25 50
1 where A = matrix of co  factors AT det C 1 = 0.179 0.967 C 0.179 0.033 23
C 1 =
0 0.179 0.967 80 25 0.186 5.388 54.6 = D= 0.179 0.033 25 50 1 0 1 84.6 To find the angles, we must calculate the unit eigenvectors:
0.186 0.183 1 0.983 0.983 0.183 R= 0.183 0.983
5.388 0.938 1 0.183
And then assemble them into a rotation matrix R so that det R = +1.
det R = (0.983)(0.983) (0.183)(0.183) = 1
The rotation matrix has the form
cos R= sin
sin cos
D = RT M R
So = 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 = 0 54.6 y54.6 MPao
100.5 84.6 MPa
84.6 MPa
10.5o
x
So, the transformation equations, Mohrs 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 = 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 x1 xy M = 68.8 4.15 = yx y1 Again, the transformation equations, Mohrs circle, and the stress tensor approach all give the same result.4.15 MPa 68.8 MPa26
y
y1 4.15 MPa
25.8 MPa
30
o
x 25.8 MPa x1