Download - Stress Transformation
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1P4 Stress and Strain Dr. A.B. ZavatskyHT08
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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2Normal 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 PNormal section
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3PV
N
P
x
y
The force P can be resolved into components:Normal force N perpendicular to the inclined plane, N = P cos Shear force V tangential to the inclined plane V = P sin
If we know the areas on which the forces act, we can calculate the associated stresses.
x
y
areaA
area A( / cos )
x
y
area
A
area A( / cos )
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4( ) 2 cos12
cos
cos cos cos
2x
2
+==
====x
AP
AP
AreaN
AreaForce
/
( ) 2 sin2
cos sin
cos sin cos sin
x
x
AP
AP
AreaV
AreaForce
==
====/
x
y
max = x occurs when = 0
max = x/2 occurs when = -/+ 45
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5Introduction 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
= x P A/xx
y
x = x P A/
P x = AP / x
y
Area A
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6Maximum stresses on a bar in tension
PPa
x = max = P / Ax
aMaximum normal stress,
Zero shear stress
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7PPab
Maximum stresses on a bar in tension
bx/2
x/2
max = x/2
= 45
Maximum shear stress,Non-zero normal stress
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8Stress 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
x x
y1x1
-
9y
z
x
Normal stresses x, y, z(tension is positive)
x x
y
y
z
xy
yx
xzzxzyyz
Shear stresses xy = yx, xz = zx, yz = zy
Sign convention for abSubscript 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 x = xx, y = yy, z = zz
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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 (z = 0 and zx = xz = zy = yz = 0).Such an element could be located on the free surface of a body (no stresses acting on the free surface).
x x
y
y
xyyx
xyyx
xy
Plane stress element in 2D
x, yxy = yxz = 0
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11
Stresses on Inclined Sections
x x
y
xyyx
xyyx
xy
x
y
x1
y1
x1
x1
y1
y1
x y1 1 y x1 1
x y1 1 y 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 or . This value may occur at some angle other than = 0. (Remember that for uni-axial tension the maximum shear stress occurred when = 45 degrees. )
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Transformation Equations
y
x1 x
y1x1
x
y
x y1 1
xyyx
Stresses
y
x1 A sec x
y1x1
xA
y A tan
x y1 1 A sec
xy A yx A tan
Forces
Left face has area A.
Bottom face has area A tan .Inclined face has area A sec .
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
x1 A sec x
y1x1
xA
y A tan
x y1 1 A sec
xy A yx A tan
( ) ( ) ( ) ( ) 0costansintansincossec :direction in the forces Sum1 1 = AAAAAx
yxyxyxx
( ) ( ) ( ) ( ) 0sintancostancossinsec :direction in the forces Sum11 1 =+ AAAAAy
yxyxyxyx
( ) ( )
2211
221
sincoscossin
cossin2sincos
:gives gsimplifyin and Using
+=++=
=
xyyxyx
xyyxx
yxxy
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22sincossin
22cos1sin
22cos1cos
identities tric trigonomefollowing theUsing
22 ==+=
2sin2cos22
:for 90 substitute face, on the stressesFor
1
1
xyyxyx
y
y
+=+
yxyx
yx +=+ 11
11 :gives and for sexpression theSumming
( )
2cos2sin2
2sin2cos22
:stress planefor equationsation transform thegives
11
1
xyyx
yx
xyyxyx
x
+=
+++=HLT, page 108
Can be used to find y1, 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:x = -80 MPa y = 50 MPaxy = -25 MPaWe need to find x1, y1, and x1y1 when = -30.
Substitute numerical values into the transformation equations:
( ) ( ) ( ) MPa9.25302sin25302cos2
50802
5080
2sin2cos22
1
1
=+++=
+++=
x
xyyxyx
x
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( )( ) ( ) ( ) ( ) MPa8.68302cos25302sin
25080
2cos2sin2
11
11
=+=
+=
yx
xyyx
yx
( ) ( ) ( ) MPa15.4302sin25302cos
25080
25080
2sin2cos22
1
1
=+=
+=
y
xyyxyx
y
Note that y1 could also be obtained (a) by substituting +60 into the equation for x1 or (b) by using the equation x + y = x1 + y1+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 (1 and 2) are known as the principal stresses. To find the principal stresses, we must differentiate the transformation equations.
( ) ( )( )
yx
xyp
xyyxx
xyyxx
xyyxyx
x
dddd
=
=+=
=+=
+++=
22tan
02cos22sin
02cos22sin22
2sin2cos22
1
1
1
p are principal angles associated with the principal stresses (HLT, page 108)
There are two values of 2p in the range 0-360, with values differing by 180.There are two values of p 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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2sin2cos22
22tan
1 xyyxyx
x
yx
xyp
+++==
We can now solve for the principal stresses by substituting for pin the stress transformation equation for x1. This tells us which principal stress is associated with which principal angle.
2p
xy
(x y) / 2
R
R
R
R
xyp
yxp
xyyx
=
=
+
=
2sin
22cos
22
22
+
++=
RRxy
xyyxyxyx
2221
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Substituting for R and re-arranging gives the larger of the two principal stresses:
22
1 22 xyyxyx +
++=
To find the smaller principal stress, use 1 + 2 = x + y.
22
12 22 xyyxyx
yx +
+=+=
These equations can be combined to give:
22
2,1 22 xyyxyx +
+= Principal stresses(HLT page 108)
To find out which principal stress goes with which principal angle, we could use the equations for sin p and cos p or for x1.
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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 2pHence, the shear stresses are zero on the principal planes.
( )( )( ) 02cos22sin
02cos22sin
02cos2sin2
0 and 0for equations theCompare
1
11
111
=+==+
=+===
xyyxx
xyyx
xyyx
yx
xyx
dd
dd
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Principal Stresses
22
2,1 22 xyyxyx +
+=
yx
xyp
=2
2tan
Principal Angles defining the Principal Planes
x
y
1 p1
1
2
2
p2
y
x x
y
xyyx
xyyx
xy
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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:x = -80 MPa y = 50 MPaxy = -25 MPa
( )MPa6.84MPa6.54
6.6915252
50802
5080
22
21
22
2,1
22
2,1
===+
+=
+
+=
xyyxyx
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( )
=+=
===
5.100,5.10
18021.0and0.212
3846.05080
2522tan
22tan
p
p
p
yx
xyp
( ) ( ) ( ) MPa6.845.102sin255.102cos2
50802
5080
2sin2cos22
1
1
=+++=
+++=
x
xyyxyx
x
But we must check which angle goes with which principal stress.
1 = 54.6 MPa with p1 = 100.52 = -84.6 MPa with p2 = 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
xxx
y
y
x, y, xy = yx =
yp
zp
xp1
1
2
2
3 = 0
1, 2, 3 = 0Usually we take 1 > 2 > 3. Since principal stresses can be com-pressive as well as tensile, 3 could be a negative (compressive) stress, rather than the zero stress.
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Maximum Shear Stress
( )( )
=
==
+=
xy
yxs
xyyxyx
xyyx
yx
dd
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 2s in the range 0-360, with values differing by 180.There are two values of s 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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2sxy
(
x
y
)
/
2
R
( )
2cos2sin2
22tan
11 xyyx
yx
xy
yxs
+=
=
We can now solve for the maximum shear stress by substituting for s in the stress transformation equation for x1y1.
R
R
R
yxs
xys
xyyx
22sin
2cos
22
22
=
=
+
=
maxmin2
2
max 2 =+
= xyyx
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Use equations for sin s and cos s or x1y1 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 s in the equations for x1 and y1 to get
syx
yx =+== 211
x
y
s s
s
s
s
max
max
max
maxy
x x
y
xyyx
xyyx
xy
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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:x = -80 MPa y = 50 MPaxy = -25 MPa
( ) MPa6.69252
5080
2
22
max
22
max
=+
=
+
=
xyyx
MPa152
50802
=+=
+=
s
yxs
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( )=
+==
=
=
5.55,5.341800.69and0.692
6.2252
50802
2tan
s
s
xy
yxs
( )( ) ( ) ( ) ( ) MPa6.695.342cos255.342sin
25080
2cos2sin2
11
11
=+=
+=
yx
xyyx
yx
But we must check which angle goes with which shear stress.
max = 69.6 MPa with smax = 55.5min = -69.6 MPa with smin = -34.5
15 MPa
15 MPa
15 MPa
x
y
15 MPa
69.6 MPa
-34.5o55.5o
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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
==
+
=
+
+=
xyyx
xyyxyx
pp
s
yx
xyp
xy
yxs
2cot2tan12tan
22tan
22tan
===
=
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( )
==
==
=+=+
=+
45
45
9022
022cos
02cos2cos2sin2sin
02sin2cos
2cos2sin
02cot2tan
ps
ps
ps
ps
psps
p
p
s
s
ps
So, the planes of maximum shear stress (s) occur at 45 to the principal planes (p).
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80 MPa 80 MPa
50 MPa
xy
50 MPa
25 MPa
Original Problem
x = -80, y = 50, xy = 25
84.6 MPa
84.6 MPa
54.6 MPa
x
y
54.6 MPa
10.5o
100.5o
Principal Stresses
1 = 54.6, 2 = 0, 3 = -84.6
15 MPa
15 MPa
15 MPa
x
y
15 MPa
69.6 MPa
-34.5o55.5o
Maximum Shear
max = 69.6, s = -15
==
=
5.55,5.34455.10
45
s
s
ps
( )
MPa6.692
6.846.542
max
max
21max
=
=
=