Chapter 6 Transformation
Transcript of Chapter 6 Transformation
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Chapter 6
.
Materials for this chapter are Originally taken from :
Ferdinand P. Beer, E Russell Johnston, Jr, John T. Dewolf and David Mazurek: Mechanics of Materials ,th
FOR INTERNAL USE ONLY
Introduction
• The most general state of stress at a point may
be re resented b 6 com onents,
stressesshearing,,
stressesnormal,,
zx yz xy
z y x
τ τ τ
σ σ σ
),, :(Note xz zx zy yz yx xy τ τ τ τ τ τ ===
• ame state o stress s represente y a
different set of components if axes are rotated.
• The first part of the chapter is concerned with
how the components of stress are transformed
under a rotation of the coordinate axes. The
second part of the chapter is devoted to a
components of strain.
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Introduction
• Plane Stress - state of stress in which two faces of
t e cu c e ement are ree o stress. For t e
illustrated example, the state of stress is defined by
.,, xy zy zx z y x
• State of plane stress occurs in a thin plate subjected
to forces acting in the midplane of the plate.
• State of plane stress also occurs on the free surface
of a structural element or machine component, i.e.,
at any point of the surface not subjected to an
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.
Transformation of Plane Stress
• Consider the conditions for equilibrium of a
rismatic element with faces er endicular to
( ) ( ) θ θ τ θ θ σ σ sincoscoscos0 A A AF xy x x x Δ−Δ−Δ==∑ ′′
the x, y, and x’ axes.
( ) ( )
( ) ( ) θ θ τ θ θ σ τ
θ θ τ θ θ σ
coscossincos0
cossinsinsin
A A AF
A A
xy x y x y
xy y
Δ−Δ+Δ==∑
Δ−Δ−
′′′
( ) ( ) θ θ τ θ θ σ sinsincossin A A xy y Δ+Δ−
θ τ θ σ σ σ σ
σ 2sin2cos y x y x
+−
++
=′
• The equations may be rewritten to yield
θ τ θ σ σ σ σ
σ 2sin2cos
22
22
xy y x y x
y −−
−+
=′
θ τ θ σ σ
τ 2cos2sin2
xy y x
y x +−
−=′′
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Principal Stresses
• The previous equations are combined to
ield arametric e uations for a circle,
( ) 222 y xave x Rτ σ σ =+− ′′′
22
xy y x y x
ave R τ σ σ σ σ
σ +⎟ ⎞
⎜⎛ −
=+
=
•
planes of stress with zero shearing stresses.
2− 2
minmax,
2
22xy
y x y x
τ
τ σ +
⎟
⎟
⎠⎜
⎜
⎝
±=
o90 bse aratedan lestwodefines: Note
2tan y x
pσ σ
θ −
=
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Maximum Shearing Stress
Maximum shearing stress occurs for ave x σ σ =′
2
22
max xy y x
R τ σ σ
τ +⎟⎟ ⎞
⎜⎜⎛ −
==
22tan xy
y x
s τ
σ σ
θ
−−=
45 byfromoffset
and90 byseparatedanglestwodefines: Note
o
o
θ
2
y xave
σ σ σ σ
+==′
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Example 6.1
SOLUTION:
• n e e emen or en a on or e pr nc pa
stresses from
xτ 2
y x p σ σ −=an
• Determine the rinci al stresses from
22
minmax,22
xy y x y x
τ σ σ σ σ
σ +⎟⎟ ⎞
⎜⎜⎛ −
±+
=Fig. 7.13
,
determine (a) the principal planes,
(b) the principal stresses, (c) the• Calculate the maximum shearing stress with
maximum shearing stress and the
corresponding normal stress.
2max 2 xy
y x
τ
σ σ
τ +⎟⎟ ⎠⎜⎜⎝
−
=
2
y x σ σ σ
+=′
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Example 6.1
SOLUTION:
• n e e emen or en a on or e pr nc pa
stresses from
( )+ 4022 xτ
( )
°°=
=−−
=−
=
1.233,1.532
.1050
an
p
y x p
θ
σ σ
°°= 6.116,6.26 pθ MPa40MPa50 +=+= xy x τ σ
Fig. 7.13
• e erm ne e pr nc pa s resses rom
22
⎞⎛ −+ y x y x σ σ σ σ
MPa10−= xσ
( ) ( )22
m nmax,
403020
22
+±=
⎠⎝ xy
MPa30
MPa70max
−=
=
σ
σ
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Fig. 7.14
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Example 6.1
• Calculate the maximum shearing stress with
22
max2
+⎟⎟ ⎠
⎞⎜⎜⎝
⎛ −= xy
y xτ
σ σ τ
( ) ( )22 4030 +=
MPa50max =τ
45−= ps θ θ MPa40MPa50 +=+= xy x τ σ
Fig. 7.13
°°−= 6.71,4.18sθ MPa10−= xσ
1050 −=
+==′ y x
ave
σ σ σ σ
• e correspon ng norma s ress s
MPa20=′σ
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Fig. 7.16
Example 6.2
A plane stress system is as shown SOLUTION:
n t e agram e ow. Determ ne
the direct and shear stress acting
on the lane P-P.
fordirectandshearstress:
σ σ σ σ + −cos s n
2 2
10 2 10 2 cos2 75 8 sin 2 75
N xy
o o
σ τ = + +
− += + + −
( )
2 2
5.196 compressive MPa= −
sin 2 cos22
x y
xy
σ σ τ θ τ θ
−= − +
)sin 2 75 8 cos2 752
4.928
o o
MPa
= − + −
=
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Mohr’s Circle for Plane Stress
• With the physical significance of Mohr’s circle
for lane stress established, it ma be a lied
with simple geometric considerations. Critical
values are estimated graphically or calculated.
• For a known state of plane stress
lot the oints X and Y and construct the xy y x τ σ σ ,,
circle centered at C .
22
y x y x σ σ σ σ ⎞⎛ −+
22xyave
⎠⎝ ==
• e pr nc pa s resses are o a ne a an .
ave Rσ σ ±=minmax,
y x
xy p
σ σ θ
−=2tan
The direction of rotation of Ox to Oa is
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the same as CX to CA.
Mohr’s Circle for Plane Stress
• With Mohr’s circle uniquely defined, the state
o stress at ot er axes or entat ons may e
depicted.
• For the state of stress at an angle θ with
res ect to the xy axes, construct a new
diameter X’Y’ at an angle 2θ with respect to
XY .
• Normal and shear stresses are obtained
’ ’ .
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Mohr’s Circle for Plane Stress
• Mohr’s circle for centric axial loading:
0, === xy y x A
τ σ σ A
xy y x2
=== τ σ σ
• ’
Tc Tc
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J xy y x === τ σ σ === xy y x
J τ σ σ
Example 6.3
Fig. 7.13
,
(a) construct Mohr’s circle, determine
(b) the principal planes, (c) the SOLUTION:
principal stresses, (d) the maximum
shearing stress and the corresponding• Construction of Mohr’s circle
( ) ( )1050−++
y xσ σ
.
MPa40MPa302050
22
==−= FX CF
ave
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( ) ( ) MPa504030 22 =+== CX R
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Example 6.3
• Principal planes and stresses
5020max +=+== CAOC OAσ
MPa70max =σ
5020min −=−== BC OC OBσ
MPa30min −=σ
==40
2tan pFX
θ
°= 1.532 pθ
°= 6.26θ
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Example 6.3
• Maximum shear stress
°+= 45 ps θ θ
°= 6.71sθ
R=maxτ
MPa50max =τ
aveσ σ =′
MPa20=′σ
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Sample Problem 6.4
For the state of stress shown,
e erm ne a e pr nc pa p anes
and the principal stresses, (b) the
stress com onents exerted on the
element obtained by rotating thegiven element counterclockwise
SOLUTION:
• Construct Mohr’s circlet roug egrees.
MPa802
60100
2=
+=
+=
y xave
σ σ σ
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( ) ( ) ( ) ( ) MPa524820 =+=+= FX CF R
Sample Problem 6.4
• Principal planes and stresses
48 XF max +== CAOC OAσ max −== BC OC OAσ
°=
===
4.672
.20
an
p
pCF
θ 5280 += 5280 −=
MPa132max +=σ MPa28min +=σ
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clockwise7.33 °= pθ
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Sample Problem 6.4
°−=−==
°=°−°−°=
′ 6.52cos5280
6.524.6760180
KC OC OK xσ
φ • Stress components after rotation by 30o
’ ’ ’
°=′=
°+=+==
′′
′
6.52sin52
6.52cos5280
X K
CLOC OL
y x
y
τ
σ o n s an on o r s c rc e a
correspond to stress components on therotated element are obtained b rotatin
XY counterclockwise through °= 602θ
MPa6.111
MPa4.48
+=
+=
′
′
y
x
σ
σ
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MPa3.41=′′ y xτ
Supplementary Problem 6.1
1. At a point in the structural member, the stresses
are represented as in figure below. Find ;
(a) the magnitude and orientation of the
principal stresses and
(b) the magnitude and orientation of the
normal stresses.
2. For the state of plane stress shown, determine (a)
the principal planes and the principal stresses, (b)
the stress components exerted on the element
obtained by rotating the given element
counterclockwise via 30o.
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Supplementary Problem 6.1
3. A structural member is sub ected to two set of loadin s. Each se aratel roduce the
.
stress conditions at a point A as shown in Figure Q1(a) and FigureQ1(b).
a) Determining σx, σy and τxy for the stress conditions.
b) By continuing the effect of both stress conditions Q1 (a) and Q2(b), determine
the principal stresses and their orientations with respect to the x-axis and also
e magn u e o e max mum s ear ng s ress.
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General State of Stress
• Consider the general 3D state of stress at a point and
the transformation of stress from element rotation
• State of stress at Q defined by: zx yz xy z y x τ τ τ σ σ σ ,,,,,
• Consider tetrahedron with face perpendicular to the
line QN with direction cosines: z y x λ λ λ ,,
• The requirement leads to,∑ = 0nF
222=
x z zx z y yz y x xy
z z y y x xn
λ λ τ λ λ τ λ λ τ 222 +++
• Form of equation guarantees that an element
orientation can be found such that
ccbbaan λ σ λ σ λ σ σ ++=
These are the principal axes and principal planes
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Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress
• The three circles represent the• Transformation of stress for an element
normal and shearing stresses for
rotation around each principal axis.
rotated around a principal axis may be
represented by Mohr’s circle.
• Points A, B, and C represent the
principal stresses on the principal planes• Radius of the largest circle yields the
maximum shearing stress.
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minmaxmax
2σ σ τ −=
Application of Mohr’s Circle to the Three-Dimensional Analysis of Stress
,
perpendicular to the plane of stress is a
principal axis (shearing stress equal zero).
• If the points A and B (representing the
principal planes) are on opposite sides of
a) the corresponding principal stresses
the origin, then
are the maximum and minimum
normal stresses for the element
b) the maximum shearing stress for the
element is equal to the maximum “in-
”
c) planes of maximum shearing stresso .