Lecture Mechanics of materials (Third edition) - Chapter 7: Transformations of stress and strain
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<b>MECHANICS OF </b>
<b>MATERIALS</b>
<b>Ferdinand P. Beer</b>
<b>E. Russell Johnston, Jr.</b>
<b>John T. DeWolf</b>
<b>Lecture Notes:</b>
<b>J. Walt Oler</b>
<b>Texas Tech University</b>
CHAPTER
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Transformations of Stress and Strain
Introduction
Transformation of Plane Stress
Principal Stresses
Maximum Shearing Stress
Example 7.01
Sample Problem 7.1
Mohr’s Circle for Plane Stress
Example 7.02
Sample Problem 7.2
General State of Stress
Application of Mohr’s Circle to the Three- Dimensional Analysis of Stress
Yield Criteria for Ductile Materials Under Plane Stress
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Introduction
• The most general state of stress at a point may
be represented by 6 components,
)
,
,
:
(Note
stresses
shearing
,
,
stresses
normal
,
,
<i>xz</i>
<i>zx</i>
<i>zy</i>
<i>yz</i>
<i>yx</i>
<i>xy</i>
<i>zx</i>
<i>yz</i>
<i>xy</i>
<i>z</i>
<i>y</i>
<i>x</i>
τ
τ
τ
τ
τ
τ
τ
τ
τ
σ
σ
σ
=
=
=
• Same state of stress is represented by a
different set of components if axes are rotated.
• The first part of the chapter is concerned with
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Introduction
• <i>Plane Stress</i> - state of stress in which two faces of
the cubic element are free of stress. For the
illustrated example, the state of stress is defined by
.
0
,
, <i><sub>y</sub></i> <sub>xy</sub> and <i><sub>z</sub></i> = <i><sub>zx</sub></i> = <i><sub>zy</sub></i> =
<i>x</i> σ τ σ τ τ
σ
• State of plane stress occurs in a thin plate subjected
to forces acting in the midplane of the plate.
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Transformation of Plane Stress
(
)
(
)
(
)
(
)
(
)
(
)
(
θ)
θ τ(
θ)
θσ
θ
θ
τ
θ
θ
σ
τ
θ
θ
τ
θ
θ
σ
θ
θ
τ
θ
θ
σ
σ
sin
sin
cos
sin
cos
cos
sin
cos
0
cos
sin
sin
sin
sin
cos
cos
cos
0
<i>A</i>
<i>A</i>
<i>A</i>
<i>A</i>
<i>A</i>
<i>F</i>
<i>A</i>
<i>A</i>
<i>A</i>
<i>A</i>
<i>A</i>
<i>F</i>
<i>xy</i>
<i>y</i>
<i>xy</i>
<i>x</i>
<i>y</i>
<i>x</i>
<i>y</i>
<i>xy</i>
<i>y</i>
<i>xy</i>
<i>x</i>
<i>x</i>
<i>x</i>
∆
+
∆
−
∆
−
∆
+
∆
=
=
∑
∆
−
∆
−
∆
−
∆
−
∆
=
=
∑
′
′
′
′
′
• Consider the conditions for equilibrium of a
prismatic element with faces perpendicular to
the <i>x</i>, <i>y</i>, and <i>x’</i> axes.
θ
τ
θ
σ
σ
τ
θ
τ
θ
σ
σ
σ
σ
σ
θ
τ
θ
σ
σ
σ
σ
σ
2
cos
2
sin
2
2
sin
2
cos
2
2
2
sin
2
cos
2
2
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Principal Stresses
• The previous equations are combined to
yield parametric equations for a circle,
(
)
2
2
2
2
2
2
2
where
<i>xy</i>
<i>y</i>
<i>x</i>
<i>y</i>
<i>x</i>
<i>ave</i>
<i>y</i>
<i>x</i>
<i>ave</i>
<i>x</i>
<i>R</i>
<i>R</i>
τ
σ
σ
σ
σ
σ
τ
σ
σ
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
=
+
=
=
+
− <sub>′</sub> <sub>′</sub>
′
• <i>Principal stresses</i> occur on the <i>principal </i>
<i>planes of stress</i> with zero shearing stresses.
2
2
min
max,
2
2
tan
2
2
<i>y</i>
<i>x</i>
<i>xy</i>
<i>p</i>
<i>xy</i>
<i>y</i>
<i>x</i>
<i>y</i>
<i>x</i>
σ
σ
τ
θ
τ
σ
σ
σ
σ
σ
−
=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
±
+
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Maximum Shearing Stress
<i>Maximum shearing stress</i> occurs for σ<i>x</i>′ =σ<i>ave</i>
2
45
by
from
offset
and
90
by
separated
angles
two
defines
:
Note
2
2
tan
2
o
o
2
2
max
<i>y</i>
<i>x</i>
<i>ave</i>
<i>p</i>
<i>xy</i>
<i>y</i>
<i>x</i>
<i>s</i>
<i>xy</i>
<i>y</i>
<i>x</i>
<i>R</i>
σ
σ
σ
σ
θ
τ
σ
σ
θ
τ
σ
σ
τ
+
=
=
′
−
−
=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
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Example 7.01
For the state of plane stress shown,
determine (a) the principal panes,
(b) the principal stresses, (c) the
maximum shearing stress and the
corresponding normal stress.
SOLUTION:
• Find the element orientation for the principal
stresses from
<i>y</i>
<i>x</i>
<i>xy</i>
<i>p</i> <sub>σ</sub> <sub>σ</sub>
τ
θ
−
= 2
2
tan
• Determine the principal stresses from
2
2
min
max,
2
2 <i>xy</i>
<i>y</i>
<i>x</i>
<i>y</i>
<i>x</i> σ σ σ <sub>τ</sub>
σ
σ <sub>⎟⎟</sub> +
⎠
⎞
⎜⎜
⎝
⎛ −
±
+
=
• Calculate the maximum shearing stress with
2
2
max
2 <i>xy</i>
<i>y</i>
<i>x</i> σ <sub>τ</sub>
σ
τ <sub>⎟⎟</sub> +
⎠
⎞
⎜⎜
⎝
⎛ −
=
<i>y</i>
<i>x</i> σ
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Example 7.01
SOLUTION:
• Find the element orientation for the principal
stresses from
(
)
(
)
°
°
=
=
−
−
+
=
−
=
1
.
233
,
1
.
53
2
333
.
1
10
50
40
2
2
2
tan
<i>p</i>
<i>y</i>
<i>x</i>
<i>xy</i>
<i>p</i>
θ
σ
σ
τ
θ
°
°
= 26.6 ,116.6
<i>p</i>
θ
MPa
10
MPa
40
MPa
50
−
=
+
=
+
=
<i>x</i>
<i>xy</i>
<i>x</i>
σ
τ
σ
• Determine the principal stresses from
( ) ( )
2 22
2
min
max,
40
30
20
2
2
+
±
=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
±
+
=σ<i>x</i> σ <i>y</i> σ<i>x</i> σ<i>y</i> τ<i><sub>xy</sub></i>
σ
MPa
70
max =
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Example 7.01
MPa
10
MPa
40
MPa
50
−
=
+
=
+
=
<i>x</i>
<i>xy</i>
<i>x</i>
σ
τ
σ
• Calculate the maximum shearing stress with
( ) ( )
2 22
2
max
40
30
2
+
=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
= σ<i>x</i> σ <i>y</i> τ<i><sub>xy</sub></i>
τ
MPa
50
max =
τ
45
−
= <i><sub>p</sub></i>
<i>s</i> θ
θ
°
°
−
= 18.4 , 71.6
<i>s</i>
θ
2
10
50
2
−
=
+
=
=
′ <i>x</i> <i>y</i>
<i>ave</i>
σ
σ
σ
σ
• The corresponding normal stress is
=
′
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