Third Edition
CHAPTER
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Transformations of
Stress and Strain
Lecture Notes:
J. Walt Oler
Texas Tech University
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Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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
Fracture Criteria for Brittle Materials Under Plane Stress
Stresses in Thin-Walled Pressure Vessels
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Introduction
• The most general state of stress at a point may
be represented by 6 components,
σ x ,σ y ,σ z
normal stresses
τ xy , τ yz , τ zx shearing stresses
(Note : τ xy = τ yx , τ yz = τ zy , τ zx = τ xz )
• 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.
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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
σ x , σ y , τ xy and σ z = τ zx = τ zy = 0.
• 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
external force.
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Transformation of Plane Stress
• Consider the conditions for equilibrium of a
prismatic element with faces perpendicular to
the x, y, and x’ axes.
∑ Fx′ = 0 = σ x′∆A − σ x (∆A cosθ ) cosθ − τ xy (∆A cosθ )sin θ
− σ y (∆A sin θ )sin θ − τ xy (∆A sin θ ) cosθ
∑ Fy ′ = 0 = τ x′y ′∆A + σ x (∆A cosθ )sin θ − τ xy (∆A cosθ ) cosθ
− σ y (∆A sin θ ) cosθ + τ xy (∆A sin θ )sin θ
• The equations may be rewritten to yield
σ
σ
τ
σ
σ
σ
2
σ
σ
2
σ
σ
σ
2
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σ
2
2
σ
cos 2θ τ
sin 2θ
cos 2θ τ
sin 2θ
sin 2θ τ
cos 2θ
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Principal Stresses
• The previous equations are combined to
yield parametric equations for a circle,
(σ x′ − σ ave )2 + τ x2′y′ = R 2
where
σ ave =
2
σ x +σ y
⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
R = ⎜⎜
2
⎝
⎠
2
• Principal stresses occur on the principal
planes of stress with zero shearing stresses.
σ max,min =
tan 2θ p =
σ x +σ y
2
2
⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
± ⎜⎜
2
⎝
⎠
2τ xy
σ x −σ y
Note : defines two angles separated by 90o
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Maximum Shearing Stress
Maximum shearing stress occurs for
σ x′ = σ ave
2
⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
τ max = R = ⎜⎜
2
⎝
⎠
σ x −σ y
tan 2θ s = −
2τ xy
Note : defines two angles separated by 90o and
offset from θ p by 45o
σ ′ = σ ave =
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σ x +σ y
2
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Example 7.01
SOLUTION:
• Find the element orientation for the principal
stresses from
2τ xy
tan 2θ p =
σ x −σ y
• Determine the principal stresses from
σ max,min =
σx +σ y
2
⎛σ x − σ y ⎞
2
⎟⎟ + τ xy
± ⎜⎜
2
⎝
⎠
2
For the state of plane stress shown,
determine (a) the principal panes, • Calculate the maximum shearing stress with
(b) the principal stresses, (c) the
2
σ
σ
−
⎛
⎞
x
y
maximum shearing stress and the
2
⎟⎟ + τ xy
τ max = ⎜⎜
2
corresponding normal stress.
⎝
⎠
σx +σ y
′
σ =
2
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Example 7.01
SOLUTION:
• Find the element orientation for the principal
stresses from
2τ xy
2(+ 40 )
= 1.333
=
tan 2θ p =
σ x − σ y 50 − (− 10 )
2θ p = 53.1°, 233.1°
σ x = +50 MPa
σ x = −10 MPa
θ p = 26.6°, 116.6°
τ xy = +40 MPa
• Determine the principal stresses from
σ max,min =
σx +σ y
2
= 20 ±
2
⎛σ x − σ y ⎞
2
⎟⎟ + τ xy
± ⎜⎜
2
⎠
⎝
(30)2 + (40)2
σ max = 70 MPa
σ min = −30 MPa
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Example 7.01
• Calculate the maximum shearing stress with
2
⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
τ max = ⎜⎜
2
⎠
⎝
=
(30)2 + (40)2
τ max = 50 MPa
σ x = +50 MPa
σ x = −10 MPa
τ xy = +40 MPa
θ s = θ p − 45
θ s = −18.4°, 71.6°
• The corresponding normal stress is
σ x + σ y 50 − 10
σ ′ = σ ave =
=
2
2
σ ′ = 20 MPa
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Sample Problem 7.1
SOLUTION:
• Determine an equivalent force-couple
system at the center of the transverse
section passing through H.
• Evaluate the normal and shearing stresses
at H.
• Determine the principal planes and
calculate the principal stresses.
A single horizontal force P of 150 lb
magnitude is applied to end D of lever
ABD. Determine (a) the normal and
shearing stresses on an element at point
H having sides parallel to the x and y
axes, (b) the principal planes and
principal stresses at the point H.
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Sample Problem 7.1
SOLUTION:
• Determine an equivalent force-couple
system at the center of the transverse
section passing through H.
P = 150 lb
T = (150 lb )(18 in ) = 2.7 kip ⋅ in
M x = (150 lb )(10 in ) = 1.5 kip ⋅ in
• Evaluate the normal and shearing stresses
at H.
σy =+
(1.5 kip ⋅ in )(0.6 in )
Mc
=+
1 π (0.6 in )4
I
4
τ xy = +
(2.7 kip ⋅ in )(0.6 in )
Tc
=+
1 π (0.6 in )4
J
2
σ x = 0 σ y = +8.84 ksi τ y = +7.96 ksi
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Sample Problem 7.1
• Determine the principal planes and
calculate the principal stresses.
2τ xy
2(7.96 )
tan 2θ p =
=
= −1.8
σ x − σ y 0 − 8.84
2θ p = −61.0°,119°
θ p = −30.5°, 59.5°
σ max,min =
σx +σ y
2
2
⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
± ⎜⎜
2
⎠
⎝
2
0 + 8.84
⎛ 0 − 8.84 ⎞
2
=
± ⎜
⎟ + (7.96 )
2
⎝ 2 ⎠
σ max = +13.52 ksi
σ min = −4.68 ksi
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Mohr’s Circle for Plane Stress
• With the physical significance of Mohr’s circle
for plane stress established, it may be applied
with simple geometric considerations. Critical
values are estimated graphically or calculated.
• For a known state of plane stress σ x , σ y ,τ xy
plot the points X and Y and construct the
circle centered at C.
σ ave =
σ x +σ y
2
2
⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
R = ⎜⎜
2
⎝
⎠
• The principal stresses are obtained at A and B.
σ max,min = σ ave ± R
tan 2θ p =
2τ xy
σ x −σ y
The direction of rotation of Ox to Oa is
the same as CX to CA.
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Mohr’s Circle for Plane Stress
• With Mohr’s circle uniquely defined, the state
of stress at other axes orientations may be
depicted.
• For the state of stress at an angle θ with
respect to the xy axes, construct a new
diameter X’Y’ at an angle 2θ with respect to
XY.
• Normal and shear stresses are obtained
from the coordinates X’Y’.
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Mohr’s Circle for Plane Stress
• Mohr’s circle for centric axial loading:
σx =
P
, σ y = τ xy = 0
A
σ x = σ y = τ xy =
P
2A
• Mohr’s circle for torsional loading:
σ x = σ y = 0 τ xy =
Tc
J
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σx =σy =
Tc
τ xy = 0
J
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Example 7.02
For the state of plane stress shown,
(a) construct Mohr’s circle, determine
(b) the principal planes, (c) the
principal stresses, (d) the maximum
shearing stress and the corresponding
normal stress.
SOLUTION:
• Construction of Mohr’s circle
σ x + σ y (50 ) + (− 10 )
=
= 20 MPa
σ ave =
2
2
CF = 50 − 20 = 30 MPa FX = 40 MPa
R = CX =
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(30)2 + (40)2 = 50 MPa
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Example 7.02
• Principal planes and stresses
σ max = OA = OC + CA = 20 + 50
σ max = 70 MPa
σ max = OB = OC − BC = 20 − 50
σ max = −30 MPa
FX 40
=
CP 30
2θ p = 53.1°
tan 2θ p =
θ p = 26.6°
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Example 7.02
• Maximum shear stress
θ s = θ p + 45°
τ max = R
σ ′ = σ ave
θ s = 71.6°
τ max = 50 MPa
σ ′ = 20 MPa
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Sample Problem 7.2
For the state of stress shown,
determine (a) the principal planes
and the principal stresses, (b) the
stress components exerted on the
element obtained by rotating the SOLUTION:
given element counterclockwise • Construct Mohr’s circle
through 30 degrees.
σ x + σ y 100 + 60
=
= 80 MPa
σ ave =
2
2
R=
(CF )2 + (FX )2 = (20)2 + (48)2 = 52 MPa
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Sample Problem 7.2
• Principal planes and stresses
XF 48
=
= 2.4
CF 20
2θ p = 67.4°
tan 2θ p =
θ p = 33.7° clockwise
σ max = OA = OC + CA
= 80 + 52
σ max = +132 MPa
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σ max = OA = OC − BC
= 80 − 52
σ min = +28 MPa
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Sample Problem 7.2
• Stress components after rotation by
30o
Points X’ and Y’ on Mohr’s circle that
correspond to stress components on the
rotated element are obtained by rotating
XY counterclockwise through 2θ = 60°
φ = 180° − 60° − 67.4° = 52.6°
σ x′ = OK = OC − KC = 80 − 52 cos 52.6°
σ y′ = OL = OC + CL = 80 + 52 cos 52.6°
τ x′y′ = KX ′ = 52 sin 52.6°
σ x′ = +48.4 MPa
σ y′ = +111.6 MPa
τ x′y′ = 41.3 MPa
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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: σ x ,σ y ,σ z ,τ xy ,τ yz ,τ zx
• Consider tetrahedron with face perpendicular to the
line QN with direction cosines: λx , λ y , λz
• The requirement ∑ Fn = 0 leads to,
σ n = σ x λ2x + σ y λ2y + σ z λ2z
+ 2τ xy λx λ y + 2τ yz λ y λz + 2τ zx λz λx
• Form of equation guarantees that an element
orientation can be found such that
σ n = σ a λ2a + σ bλb2 + σ cλc2
These are the principal axes and principal planes
and the normal stresses are the principal stresses.
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Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress
• Transformation of stress for an element
rotated around a principal axis may be
represented by Mohr’s circle.
• The three circles represent the
normal and shearing stresses for
rotation around each principal axis.
• Points A, B, and C represent the
• Radius of the largest circle yields the
principal stresses on the principal planes
maximum shearing stress.
1
(shearing stress is zero)
τ max = σ max − σ min
2
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Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress
• In the case of plane stress, the axis
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
the origin, then
a) the corresponding principal stresses
are the maximum and minimum
normal stresses for the element
b) the maximum shearing stress for the
element is equal to the maximum “inplane” shearing stress
c) planes of maximum shearing stress
are at 45o to the principal planes.
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Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress
• If A and B are on the same side of the
origin (i.e., have the same sign), then
a) the circle defining σmax, σmin, and
τmax for the element is not the circle
corresponding to transformations within
the plane of stress
b) maximum shearing stress for the
element is equal to half of the
maximum stress
c) planes of maximum shearing stress are
at 45 degrees to the plane of stress
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Yield Criteria for Ductile Materials Under Plane Stress
• Failure of a machine component
subjected to uniaxial stress is directly
predicted from an equivalent tensile test
• Failure of a machine component
subjected to plane stress cannot be
directly predicted from the uniaxial state
of stress in a tensile test specimen
• It is convenient to determine the
principal stresses and to base the failure
criteria on the corresponding biaxial
stress state
• Failure criteria are based on the
mechanism of failure. Allows
comparison of the failure conditions for
a uniaxial stress test and biaxial
component loading
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Yield Criteria for Ductile Materials Under Plane Stress
Maximum shearing stress criteria:
Structural component is safe as long as the
maximum shearing stress is less than the
maximum shearing stress in a tensile test
specimen at yield, i.e.,
σ
τ max < τ Y = Y
2
For σa and σb with the same sign,
τ max =
σa
2
or
σb
σ
< Y
2
2
For σa and σb with opposite signs,
τ max =
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σa −σb
2
σ
< Y
2
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Yield Criteria for Ductile Materials Under Plane Stress
Maximum distortion energy criteria:
Structural component is safe as long as the
distortion energy per unit volume is less
than that occurring in a tensile test specimen
at yield.
ud < uY
1 2
1 2
σ a − σ aσ b + σ b2 <
σ Y − σ Y × 0 + 02
6G
6G
(
)
(
)
σ a2 − σ aσ b + σ b2 < σ Y2
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Fracture Criteria for Brittle Materials Under Plane Stress
Brittle materials fail suddenly through rupture
or fracture in a tensile test. The failure
condition is characterized by the ultimate
strength σU.
Maximum normal stress criteria:
Structural component is safe as long as the
maximum normal stress is less than the
ultimate strength of a tensile test specimen.
σ a < σU
σ b < σU
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Stresses in Thin-Walled Pressure Vessels
• Cylindrical vessel with principal stresses
σ1 = hoop stress
σ2 = longitudinal stress
• Hoop stress:
∑ Fz = 0 = σ 1(2t ∆x ) − p (2r ∆x )
σ1 =
pr
t
• Longitudinal stress:
( )
2
∑ Fx = 0 = σ 2 (2π rt ) − p π r
pr
σ2 =
2t
σ 1 = 2σ 2
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Stresses in Thin-Walled Pressure Vessels
• Points A and B correspond to hoop stress, σ1,
and longitudinal stress, σ2
• Maximum in-plane shearing stress:
1
2
τ max(in − plane) = σ 2 =
pr
4t
• Maximum out-of-plane shearing stress
corresponds to a 45o rotation of the plane
stress element around a longitudinal axis
τ max = σ 2 =
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pr
2t
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Stresses in Thin-Walled Pressure Vessels
• Spherical pressure vessel:
σ1 = σ 2 =
pr
2t
• Mohr’s circle for in-plane
transformations reduces to a point
σ = σ 1 = σ 2 = constant
τ max(in -plane) = 0
• Maximum out-of-plane shearing
stress
τ max = 12 σ 1 =
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pr
4t
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MECHANICS OF MATERIALS
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Transformation of Plane Strain
• Plane strain - deformations of the material
take place in parallel planes and are the
same in each of those planes.
• Plane strain occurs in a plate subjected
along its edges to a uniformly distributed
load and restrained from expanding or
contracting laterally by smooth, rigid and
fixed supports
components of strain :
ε x ε y γ xy
(ε z = γ zx = γ zy = 0)
• Example: Consider a long bar subjected
to uniformly distributed transverse loads.
State of plane stress exists in any
transverse section not located too close to
the ends of the bar.
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Transformation of Plane Strain
• State of strain at the point Q results in
different strain components with respect
to the xy and x’y’ reference frames.
ε (θ ) = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cosθ
ε OB = ε (45°) = 12 (ε x + ε y + γ xy )
γ xy = 2ε OB − (ε x + ε y )
• Applying the trigonometric relations
used for the transformation of stress,
εx + ε y εx − ε y
γ xy
ε x′ =
+
cos 2θ +
sin 2θ
ε y′ =
γ x′y′
2
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2
2
2
εx + ε y
εx − ε y
γ xy
2
=−
−
εx − ε y
2
2
cos 2θ −
sin 2θ +
γ xy
2
2
sin 2θ
cos 2θ
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Mohr’s Circle for Plane Strain
• The equations for the transformation of
plane strain are of the same form as the
equations for the transformation of plane
stress - Mohr’s circle techniques apply.
• Abscissa for the center C and radius R ,
ε ave =
2
εx + ε y
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
⎟⎟
⎟⎟ + ⎜⎜
R = ⎜⎜
⎝ 2 ⎠ ⎝ 2 ⎠
2
2
• Principal axes of strain and principal strains,
γ xy
tan 2θ p =
εx − ε y
ε max = ε ave + R
ε min = ε ave − R
• Maximum in-plane shearing strain,
γ max = 2 R =
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(ε x − ε y )2 + γ xy2
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Three-Dimensional Analysis of Strain
• Previously demonstrated that three principal
axes exist such that the perpendicular
element faces are free of shearing stresses.
• By Hooke’s Law, it follows that the
shearing strains are zero as well and that
the principal planes of stress are also the
principal planes of strain.
• Rotation about the principal axes may be
represented by Mohr’s circles.
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Three-Dimensional Analysis of Strain
• For the case of plane strain where the x and y
axes are in the plane of strain,
- the z axis is also a principal axis
- the corresponding principal normal strain
is represented by the point Z = 0 or the
origin.
• If the points A and B lie on opposite sides
of the origin, the maximum shearing strain
is the maximum in-plane shearing strain, D
and E.
• If the points A and B lie on the same side of
the origin, the maximum shearing strain is
out of the plane of strain and is represented
by the points D’ and E’.
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Three-Dimensional Analysis of Strain
• Consider the case of plane stress,
σ x = σa σ y = σb σ z = 0
• Corresponding normal strains,
σ νσ
εa = a − b
E
εb = −
E
νσ a
σ
+ b
E
E
ν
ν
(ε a + ε b )
ε c = − (σ a + σ b ) = −
1 −ν
E
• Strain perpendicular to the plane of stress
is not zero.
• If B is located between A and C on the
Mohr-circle diagram, the maximum
shearing strain is equal to the diameter CA.
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Measurements of Strain: Strain Rosette
• Strain gages indicate normal strain through
changes in resistance.
• With a 45o rosette, εx and εy are measured
directly. γxy is obtained indirectly with,
γ xy = 2ε OB − (ε x + ε y )
• Normal and shearing strains may be
obtained from normal strains in any three
directions,
ε1 = ε x cos 2 θ1 + ε y sin 2 θ1 + γ xy sin θ1 cosθ1
ε 2 = ε x cos 2 θ 2 + ε y sin 2 θ 2 + γ xy sin θ 2 cosθ 2
ε 3 = ε x cos 2 θ 3 + ε y sin 2 θ 3 + γ xy sin θ 3 cosθ 3
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