lecture 5 plane stress transformation equations
TRANSCRIPT
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Lecture 5
Plane Stress Transformation Equations
Stress elements and plane stress. Stresses on inclined sections. Transformation equations. Principal stresses, angles, and planes.Maximum shear stress.
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Normal and shear stresses on inclined sections
To obtain a complete picture of the stresses in a bar, we must consider the stresses acting on an “inclined” (as opposed to a “normal”) section through the bar.
PP
Normal sectionInclined section
Because the stresses are the same throughout the entire bar, thestresses on the sections are uniformly distributed.
P Inclined section P Normal
section
3
PV
N
P
θ
x
y
The force P can be resolved into components:Normal force N perpendicular to the inclined plane, N = P cos θShear force V tangential to the inclined plane V = P sin θ
If we know the areas on which the forces act, we can calculate the associated stresses.
x
y
areaA
area A( / cos )θ
σθ
x
y
τθ
areaA
area A( / cos )θ
4
( ) 2 cos12
cos
cos cos cos
2x
2
θσθσσ
θθθσ
θ
θ
+==
====
x
AP
AP
AreaN
AreaForce
/
( ) 2 sin2
cos sin
cos sin cos sin
x θσθθστ
θθθθτ
θ
θ
x
AP
AP
AreaV
AreaForce
−=−=
−=−
=−
==/
x
yτθ
σθ
θ
σmax = σx occurs when θ = 0°
τmax = ± σx/2 occurs when θ = -/+ 45°
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Introduction to stress elementsStress elements are a useful way to represent stresses acting at some point on a body. Isolate a small element and show stresses acting on all faces. Dimensions are “infinitesimal”, but are drawn to a large scale.
x
y
z
σ = x P A/σx
x
y
σx σ = x P A/
P σx = AP / x
y
Area A
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Maximum stresses on a bar in tension
PPa
σx = σmax = P / Aσx
aMaximum normal stress,
Zero shear stress
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PPab
Maximum stresses on a bar in tension
bσx/2
σx/2
τmax = σx/2
θ = 45°
Maximum shear stress,Non-zero normal stress
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Stress Elements and Plane Stress
When working with stress elements, keep in mind that only one intrinsic state of stress exists at a point in a stressed body, regardless of the orientation of the element used to portray the state of stress.
We are really just rotating axes to represent stresses in a new coordinate system.
x
y
σx σxθ
y1
x1
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y
z
x
Normal stresses σx, σy, σz(tension is positive)
σx σx
σy
σy
σz
τxy
τyx
τxzτzx
τzy
τyz
Shear stresses τxy = τyx, τxz = τzx, τyz = τzy
Sign convention for τabSubscript a indicates the “face” on which the stress acts (positive x “face” is perpendicular to the positive x direction)Subscript b indicates the direction in which the stress actsStrictly σx = σxx, σy = σyy, σz = σzz
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When an element is in plane stress in the xy plane, only the x and y faces are subjected to stresses (σz = 0 and τzx = τxz = τzy = τyz = 0).
Such an element could be located on the free surface of a body (no stresses acting on the free surface).
σx σx
σy
σy
τxy
τyx
τxy
τyx
xy
Plane stress element in 2D
σx, σyτxy = τyxσz = 0
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Stresses on Inclined Sections
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
The stress system is known in terms of coordinate system xy. We want to find the stresses in terms of the rotated coordinate system x1y1.
Why? A material may yield or fail at the maximum value of σ or τ. This value may occur at some angle other than θ = 0. (Remember that for uni-axial tension the maximum shear stress occurred when θ = 45 degrees. )
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Transformation Equations
y
σx1
xθ
y1
x1
σx
σy
τx y1 1
τxy
τyx
θ
Stresses
y
σ θx1 A sec x
θ
y1
x1
σxA
σ θy A tan
τ θx y1 1 A sec
τxy Aτ θyx A tan
θ
Forces
Left face has area A.
Bottom face has area A tan θ.
Inclined face has area A sec θ.
Forces can be found from stresses if the area on which the stresses act is known. Force components can then be summed.
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y
σ θx1 A sec x
θ
y1
x1
σxA
σ θy A tan
τ θx y1 1 A sec
τxy Aτ θyx A tan
θ
( ) ( ) ( ) ( ) 0costansintansincossec:direction in the forces Sum
1
1
=−−−− θθτθθσθτθσθ AAAAAσx
yxyxyxx
( ) ( ) ( ) ( ) 0sintancostancossinsec:direction in the forces Sum
11
1
=−−−+ θθτθθσθτθσθτ AAAAAy
yxyxyxyx
( ) ( )θθτθθσστ
θθτθσθσσ
ττ
2211
221
sincoscossin
cossin2sincos
:gives gsimplifyin and Using
−+−−=
++=
=
xyyxyx
xyyxx
yxxy
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22sincossin
22cos1sin
22cos1cos
identities tric trigonomefollowing theUsing
22 θθθθθθθ =−
=+
=
θτθσσσσ
σ
θθ
2sin2cos22
:for 90 substitute face, on the stressesFor
1
1
xyyxyx
y
y
−−
−+
=
+ °
yxyx
yxσσσσ +=+ 11
11 :gives and for sexpression theSumming
( )θτθ
σστ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
:stress planefor equationsation transform thegives
11
1
xyyx
yx
xyyxyx
x
+−
−=
+−
++
=
Can be used to find σy1, instead of eqn above.
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Example: The state of plane stress at a point is represented by the stress element below. Determine the stresses acting on an element oriented 30° clockwise with respect to the original element.
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
Define the stresses in terms of the established sign convention:σx = -80 MPa σy = 50 MPa
τxy = -25 MPa
We need to find σx1, σy1, and τx1y1 when θ = -30°.
Substitute numerical values into the transformation equations:
( ) ( ) ( ) MPa9.25302sin25302cos2
50802
5080
2sin2cos22
1
1
−=°−−+°−−−
++−
=
+−
++
=
x
xyyxyx
x
σ
θτθσσσσ
σ
16
( )
( ) ( ) ( ) ( ) MPa8.68302cos25302sin2
5080
2cos2sin2
11
11
−=°−−+°−−−
−=
+−
−=
yx
xyyx
yx
τ
θτθσσ
τ
( ) ( ) ( ) MPa15.4302sin25302cos2
50802
5080
2sin2cos22
1
1
−=°−−−°−−−
−+−
=
−−
−+
=
y
xyyxyx
y
σ
θτθσσσσ
σ
Note that σy1 could also be obtained (a) by substituting +60° into the equation for σx1 or (b) by using the equation σx + σy = σx1 + σy1
+60°
25.8 MPa
25.8 MPa
4.15 MPa
x
y
4.15 MPa68.8 MPa
x1
y1
-30o
(from Hibbeler, Ex. 15.2)
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Principal StressesThe maximum and minimum normal stresses (σ1 and σ2) are known as the principal stresses. To find the principal stresses, we must differentiate the transformation equations.
( ) ( )
( )
yx
xyp
xyyxx
xyyxx
xyyxyx
x
dddd
σστ
θ
θτθσσθ
σ
θτθσσ
θσ
θτθσσσσ
σ
−=
=+−−=
=+−−
=
+−
++
=
22tan
02cos22sin
02cos22sin22
2sin2cos22
1
1
1
θp are principal angles associated with the principal stresses
There are two values of 2θp in the range 0-360°, with values differing by 180°.There are two values of θp in the range 0-180°, with values differing by 90°.So, the planes on which the principal stresses act are mutually perpendicular.
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θτθσσσσ
σ
σστ
θ
2sin2cos22
22tan
1 xyyxyx
x
yx
xyp
+−
++
=
−=
We can now solve for the principal stresses by substituting for θpin the stress transformation equation for σx1. This tells us which principal stress is associated with which principal angle.
2θp
τxy
(σx – σy) / 2
R
R
R
R
xyp
yxp
xyyx
τθ
σσθ
τσσ
=
−=
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
2sin
22cos
22
22
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −−+
+=
RRxy
xyyxyxyx τ
τσσσσσσ
σ2221
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Substituting for R and re-arranging gives the larger of the two principal stresses:
22
1 22 xyyxyx τ
σσσσσ +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −+
+=
To find the smaller principal stress, use σ1 + σ2 = σx + σy.
22
12 22 xyyxyx
yx τσσσσ
σσσσ +⎟⎟⎠
⎞⎜⎜⎝
⎛ −−
+=−+=
These equations can be combined to give:
22
2,1 22 xyyxyx τ
σσσσσ +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −±
+=
Principal stresses
To find out which principal stress goes with which principal angle, we could use the equations for sin θp and cos θp or for σx1.
20
The planes on which the principal stresses act are called the principal planes. What shear stresses act on the principal planes?
Solving either equation gives the same expression for tan 2θp
Hence, the shear stresses are zero on the principal planes.
( )
( )( ) 02cos22sin
02cos22sin
02cos2sin2
0 and 0for equations theCompare
1
11
111
=+−−=
=+−−
=+−
−=
==
θτθσσθ
σ
θτθσσ
θτθσσ
τ
θστ
xyyxx
xyyx
xyyx
yx
xyx
dd
dd
21
Principal Stresses
22
2,1 22 xyyxyx τ
σσσσσ +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −±
+=
yx
xyp σσ
τθ
−=
22tan
Principal Angles defining the Principal Planes
x
y
σ1
θp1
σ1
σ2
σ2
θp2
σy
σx σx
σy
τxy
τyx
τxy
τyx
xy
22
Example: The state of plane stress at a point is represented by the stress element below. Determine the principal stresses and draw the corresponding stress element.
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
Define the stresses in terms of the established sign convention:σx = -80 MPa σy = 50 MPa
τxy = -25 MPa
( )
MPa6.84MPa6.54
6.6915252
50802
5080
22
21
22
2,1
22
2,1
−==
±−=−+⎟⎠⎞
⎜⎝⎛ −−
±+−
=
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −±
+=
σσ
σ
τσσσσ
σ xyyxyx
23
( )
°°=
°+°=
=−−
−=
−=
5.100,5.10
18021.0and0.212
3846.05080
2522tan
22tan
p
p
p
yx
xyp
θ
θ
θ
σστ
θ
( ) ( ) ( ) MPa6.845.102sin255.102cos2
50802
5080
2sin2cos22
1
1
−=°−+°−−
++−
=
+−
++
=
x
xyyxyx
x
σ
θτθσσσσ
σ
But we must check which angle goes with which principal stress.
σ1 = 54.6 MPa with θp1 = 100.5°σ2 = -84.6 MPa with θp2 = 10.5°
84.6 MPa
84.6 MPa
54.6 MPa
x
y
54.6 MPa
10.5o
100.5o
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The two principal stresses determined so far are the principal stresses in the xy plane. But … remember that the stress element is 3D, so there are always three principal stresses.
y
xσxσx
σy
σy
ττ
ττ
σx, σy, τxy = τyx = τ
yp
zp
xpσ1
σ1
σ2
σ2
σ3 = 0
σ1, σ2, σ3 = 0
Usually we take σ1 > σ2 > σ3. Since principal stresses can be com-pressive as well as tensile, σ3 could be a negative (compressive) stress, rather than the zero stress.
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Maximum Shear Stress
( )
( )
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−=
=−−−=
+−
−=
xy
yxs
xyyxyx
xyyx
yx
dd
τσσ
θ
θτθσσθ
τ
θτθσσ
τ
22tan
02sin22cos
2cos2sin2
11
11
To find the maximum shear stress, we must differentiate the trans-formation equation for shear.
There are two values of 2θs in the range 0-360°, with values differing by 180°.There are two values of θs in the range 0-180°, with values differing by 90°.So, the planes on which the maximum shear stresses act are mutually perpendicular.
Because shear stresses on perpendicular planes have equal magnitudes, the maximum positive and negative shear stresses differ only in sign.
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2θs
τxy
(σx
–σ y
) / 2
R
( )θτθ
σστ
τσσ
θ
2cos2sin2
22tan
11 xyyx
yx
xy
yxs
+−
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−=
We can now solve for the maximum shear stress by substituting for θs in the stress transformation equation for τx1y1.
R
R
R
yxs
xys
xyyx
22sin
2cos
22
22
σσθ
τθ
τσσ
−−=
=
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
maxmin2
2
max 2τττ
σστ −=+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −= xy
yx
27
Use equations for sin θs and cos θs or τx1y1 to find out which face has the positive shear stress and which the negative.
What normal stresses act on the planes with maximum shear stress? Substitute for θs in the equations for σx1 and σy1 to get
syx
yx σσσ
σσ =+
==211
x
y
σs
θs
σs
σs
σs
τmax
τmax
τmax
τmax
σy
σx σx
σy
τxy
τyx
τxy
τyx
xy
28
Example: The state of plane stress at a point is represented by the stress element below. Determine the maximum shear stresses and draw the corresponding stress element.
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
Define the stresses in terms of the established sign convention:σx = -80 MPa σy = 50 MPa
τxy = -25 MPa
( ) MPa6.69252
5080
2
22
max
22
max
=−+⎟⎠⎞
⎜⎝⎛ −−
=
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
τ
τσσ
τ xyyx
MPa152
50802
−=+−
=
+=
s
yxs
σ
σσσ
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( )
°°−=°+−°−=
−=⎟⎟⎠
⎞⎜⎜⎝
⎛−
−−−=⎟
⎟⎠
⎞⎜⎜⎝
⎛ −−=
5.55,5.341800.69and0.692
6.2252
50802
2tan
s
s
xy
yxs
θθ
τσσ
θ
( )
( ) ( ) ( ) ( ) MPa6.695.342cos255.342sin2
5080
2cos2sin2
11
11
−=−−+−−−
−=
+−
−=
yx
xyyx
yx
τ
θτθσσ
τ
But we must check which angle goes with which shear stress.
τmax = 69.6 MPa with θsmax = 55.5°τmin = -69.6 MPa with θsmin = -34.5°
15 MPa
15 MPa
15 MPa
x
y
15 MPa
69.6 MPa
-34.5o
55.5o
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Finally, we can ask how the principal stresses and maximum shear stresses are related and how the principal angles and maximum shear angles are related.
2
2
22
22
21max
max21
22
21
22
2,1
σστ
τσσ
τσσ
σσ
τσσσσ
σ
−=
=−
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −=−
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −±
+=
xyyx
xyyxyx
pp
s
yx
xyp
xy
yxs
θθ
θ
σστ
θτ
σσθ
2cot2tan12tan
22tan
22tan
−=−
=
−=⎟
⎟⎠
⎞⎜⎜⎝
⎛ −−=
31
( )
°±=
°±=−
°±=−
=−
=+
=+
=+
45
45
9022
022cos
02cos2cos2sin2sin
02sin2cos
2cos2sin
02cot2tan
ps
ps
ps
ps
psps
p
p
s
s
ps
θθ
θθ
θθ
θθ
θθθθ
θθ
θθ
θθ
So, the planes of maximum shear stress (θs) occur at 45° to the principal planes (θp).
32
80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
Original Problem
σx = -80, σy = 50, τxy = 25
84.6 MPa
84.6 MPa
54.6 MPa
x
y
54.6 MPa
10.5o
100.5o
Principal Stresses
σ1 = 54.6, σ2 = 0, σ3 = -84.6
15 MPa
15 MPa
15 MPa
x
y
15 MPa
69.6 MPa
-34.5o
55.5o
Maximum Shear
τmax = 69.6, σs = -15
°°−=°±°=
°±=
5.55,5.34455.10
45
s
s
ps
θθ
θθ
( )
MPa6.692
6.846.542
max
max
21max
=
−−=
−=
τ
τ
σστ
1
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 re-write 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 (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.
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)
7
σx σx
σy
τxy
τyx
τxy
τyx
xy
σx1
τx1y1
σavg
cR
A (θ=0)
σx
τxy
AB (θ=90)
σy
-τxy
B
8
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
10
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
11
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.
12
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 diagonaliza-tion 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
factors-co 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.