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Application of Mohr’s Circle to the Three- Dimensional Analysis of Stress Yield Criteria for Ductile Materials Under Plane Stress. Fracture Criteria for Brittle Materials Under Plane St[r]
(1)MECHANICS OF MATERIALS
Ferdinand P Beer
E Russell Johnston, Jr. John T DeWolf
Lecture Notes: J Walt Oler
Texas Tech University
CHAPTER
(2)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
(3)Introduction
• The most general state of stress at a point may be represented by components,
) ,
,
: (Note
stresses shearing
, ,
stresses normal
, ,
xz zx
zy yz
yx xy
zx yz xy
z y x
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
(4)Introduction
• Plane Stress - 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 ,
, y xy and z = zx = zy =
x 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
(5)Transformation of Plane Stress ( ) ( ) ( ) ( ) ( ) ( ) ( θ) θ τ ( θ) θ σ 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 cos sin sin sin sin cos cos cos A A A A A F A A A A A F xy y xy x y x y xy y xy x x x ∆ + ∆ − ∆ − ∆ + ∆ = = ∑ ∆ − ∆ − ∆ − ∆ − ∆ = = ∑ ′ ′ ′ ′ ′
• Consider the conditions for equilibrium of a prismatic element with faces perpendicular to the x, y, and x’ axes
θ τ θ σ σ τ θ τ θ σ σ σ σ σ θ τ θ σ σ σ σ σ cos sin 2 sin cos 2 sin cos 2
(6)Principal Stresses
• The previous equations are combined to yield parametric equations for a circle,
( )
2 2
2
2
where
xy y
x y
x ave
y x ave
x
R R
τ σ
σ σ
σ σ
τ σ
σ
+ ⎟⎟ ⎠ ⎞ ⎜⎜
⎝
⎛ −
= +
=
= +
− ′ ′
′
• Principal stresses occur on the principal planes of stress with zero shearing stresses
2
max,
2
tan
2
y x
xy p
xy y
x y
x
σ σ
τ θ
τ σ
σ σ
σ σ
− =
+ ⎟⎟ ⎠ ⎞ ⎜⎜
⎝
⎛ −
± +
(7)Maximum Shearing Stress
Maximum shearing stress occurs for σx′ =σave
2
45 by from
offset
and 90
by separated angles
two defines :
Note
2
tan
2
o
o
2 max
y x
ave
p xy
y x
s
xy y
x
R
σ σ
σ σ
θ τ
σ σ
θ
τ σ
σ τ
+ =
= ′
− −
=
+ ⎟⎟ ⎠ ⎞ ⎜⎜
⎝
⎛ −
(8)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
y x
xy
p σ σ
τ θ
− =
2 tan
• Determine the principal stresses from
2
min max,
2
2 xy
y x
y
x σ σ σ τ
σ
σ ⎟⎟ +
⎠ ⎞ ⎜⎜
⎝
⎛ −
± +
=
• Calculate the maximum shearing stress with
2
max
2 xy
y
x σ τ
σ
τ ⎟⎟ +
⎠ ⎞ ⎜⎜
⎝
⎛ −
=
y
x σ
(9)Example 7.01
SOLUTION:
• Find the element orientation for the principal stresses from ( ) ( ) ° ° = = − − + = − = 233 , 53 333 10 50 40 2 tan p y x xy p θ σ σ τ θ ° °
= 26.6 ,116.6
p θ MPa 10 MPa 40 MPa 50 − = + = + = x xy x σ τ σ
• Determine the principal stresses from
( ) ( )2
2 max, 40 30 20 2 + ± = + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ± +
=σx σ y σx σy τxy σ
MPa 70
max =
(10)Example 7.01
MPa 10
MPa 40
MPa 50
− =
+ = +
=
x
xy x
σ
τ σ
• Calculate the maximum shearing stress with
( ) ( )2 2
max
40 30
2
+ =
+ ⎟⎟ ⎠ ⎞ ⎜⎜
⎝
⎛ −
= σx σ y τxy τ
MPa 50
max =
τ
45 −
= p
s θ
θ
° °
−
= 18.4 , 71.6
s
θ
2 10 50
− =
+ =
=
′ x y
ave
σ σ
σ σ
• The corresponding normal stress is
= ′