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]
Trang 1MECHANICS OF MATERIALS
Ferdinand P Beer
E Russell Johnston, Jr.
John T DeWolf
Lecture Notes:
J Walt Oler Texas Tech University
CHAPTER
Transformations of
Stress and Strain
Trang 2Introduction 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
Fracture Criteria for Brittle Materials Under Plane Stress Stresses in Thin-Walled Pressure Vessels
Trang 3• The most general state of stress at a point may
be represented by 6 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 how the components of stress are transformed under a rotation of the coordinate axes The second part of the chapter is devoted to a similar analysis of the transformation of the components of strain
Trang 4• 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 ,
σ
• 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
Trang 5( ) ( )
( 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
0
cos sin
sin sin
sin cos
cos cos
0
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
2 cos 2
sin
2 sin 2
cos 2
2
2 sin 2
cos 2
2
• The equations may be rewritten to yield
Trang 6• The previous equations are combined to yield parametric equations for a circle,
2 2
2 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 2
min max,
2 2
tan
2 2
y x
xy p
xy y
x y
x
σ σ
τ θ
τ σ
σ σ
σ σ
−
=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
± +
=
Trang 7Maximum shearing stress occurs for σx′ = σave
2
45
by from
offset
and 90
by separated angles
two defines :
Note
2 2
tan
2
o
o
2 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
+
=
=
′
−
−
=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
=
=
Trang 8For 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
τ θ
−
= 2 2
tan
• Determine the principal stresses from
2
2 min
max,
2
y x
y
σ
⎠
⎞
⎜⎜
⎝
⎛ −
±
+
=
• Calculate the maximum shearing stress with
2
2 max
y
σ
⎠
⎞
⎜⎜
⎝
⎛ −
=
y
σ
σ′ = +
Trang 9• Find the element orientation for the principal stresses from
( ) ( )
°
°
=
=
−
−
+
=
−
=
1 233 , 1 53 2
333
1 10 50
40 2
2 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
2
2 min
max,
40 30
20
2 2
+
±
=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
±
+
=σx σ y σx σy τxy σ
MPa 70
max =
σ
Trang 10MPa 10
MPa 40
MPa 50
−
=
+
= +
=
x
xy x
σ
τ σ
• Calculate the maximum shearing stress with
( ) ( )2 2
2
2 max
40 30
2 +
=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
= σx σ y τxy τ
MPa 50
max =
τ
45
−
θ
°
°
−
= 18 4 , 71 6
s
θ
2
10 50 2
−
=
+
=
=
ave
σ
σ σ
σ
• The corresponding normal stress is
MPa 20
=
′
σ