4_pure_bending.pptx Material de Mecanica dos Materiais

Third Edition
MECHANICS
MATERIALS
OF
CHAPTER
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Pure Bending
Lecture Notes:
J. Walt Oler
Texas Tech University
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Beer • Johnston • DeWolf
Pure Bending
Pure Bending Example 4.03
Other Loading Types Reinforced Concrete Beams
Symmetric Member in Pure Bending Sample Problem 4.4
Bending Deformations Stress Concentrations
Strain Due to Bending Plastic Deformations
Beam Section Properties Members Made of an Elastoplastic Material
Properties of American Standard Shapes Plastic Deformations of Members With a Single
Plane of S...
Deformations in a Transverse Cross Section
Residual Stresses
Sample Problem 4.2
Example 4.05, 4.06
Bending of Members Made of Several
Materials
Example 4.03
Eccentric Axial Loading in a Plane of Symmetry
Example 4.07
Reinforced Concrete Beams Sample Problem 4.8
Sample Problem 4.4 Unsymmetric Bending
Stress Concentrations Example 4.08
Plastic Deformations General Case of Eccentric Axial Loading
Members Made of an Elastoplastic Material
4 - 2
Third
Edition

Pure Bending
Pure Bending: Prismatic members
subjected to equal and opposite couples
acting in the same longitudinal plane
4 - 3
Third
Edition

Other Loading Types
• Eccentric Loading: Axial loading which
does not pass through section centroid
produces internal forces equivalent to an
axial force and a couple
• Transverse Loading: Concentrated or
distributed transverse load produces
internal forces equivalent to a shear
force and a couple
• Principle of Superposition: The normal
stress due to pure bending may be
combined with the normal stress due to
axial loading and shear stress due to
shear loading to find the complete state
of stress.
4 - 4
Third
Edition

Symmetric Member in Pure Bending
• Internal forces in any cross section are equivalent
to a couple. The moment of the couple is the
section bending moment.
From statics, a couple M consists of two equal
and opposite forces.
The sum of the components of the forces in any
direction is zero.
The moment is the same about any axis
perpendicular to the plane of the couple and
zero about any axis contained in the plane.
These requirements may be applied to the sums
of the components and moments of the statically
indeterminate elementary internal forces.
Fx = ∫σx dA = 0
M y = ∫ zσx dA = 0
M z = ∫− yσ x dA = M
•
•
•
•
4 - 5
Third
Edition

Bending Deformations
Beam with a plane of symmetry in pure
bending:
•
•
•
member remains symmetric
bends uniformly to form a circular arc
cross-sectional plane passes through arc center
and remains planar
length of top decreases and length of bottom
increases
a neutral surface must exist that is parallel to the
upper and lower surfaces and for which the length
does not change
stresses and strains are negative (compressive)
above the neutral plane and positive (tension)
below it
•
•
•
4 - 6
Third
Edition

Strain Due to Bending
Consider a beam segment of length L.
After deformation, the length of the neutral
surface remains L. At other sections,
L′ = (ρ − y)θ
δ = L − L′ = (ρ − y)θ − ρθ = −yθ
δ −
yθ
− y
ε =
=
= (strain varies linearly)
x
ρθ ρ
c
L
c
ε = or ρ =
m
ρ εm
y
εx = − c εm
4 - 7
Third
Edition

Stress Due to Bending
• For a linearly elastic material,
− y
σ = Eε ε
= E
x x m
c
y
= − σm (stress varies linearly)
c
• For static equilibrium,
• For static equilibrium,
y
∫σx dA = ∫− c σm
Fx = 0 =
σm
dA y
⎛
−
⎞
σ σ
M = −
y
dA = −
y⎜
dA
∫ ∫ ⎟
x m
⎠
c
⎝
σmI
0 = − y dA
∫ σm 2
c = y dA =
M ∫
c c
First moment with respect to neutral
plane is zero. Therefore, the neutral
surface must pass through the
section centroid.
Mc M
σm = =
I S
−
y
σ
Substituting σ =
x m
c
My
σ x = −
I
4 - 8
Third
Edition

Beam Section Properties
• The maximum normal stress due to bending,
Mc M
σm = =
I S
I = section moment of inertia
I
S = = section modulus
c
A beam section with a larger section modulus
will have a lower maximum stress
Consider a rectangular beam cross section,
•
3
1
12
bh
I 3
1 1
S =
=
c
= 6 bh =
Ah
6
h 2
Between two beams with the same cross
sectional area, the beam with the greater depth
will be more effective in resisting bending.
Structural steel beams are designed to have a
large section modulus.
•
4 - 9
Third
Edition

Properties of American Standard Shapes
755
Append
ix C. Properties of Rolled-Steel Shapes
(SI Units)
S Shapes
(American Standard Shapes)
106 mm4 1<>3
mm3
mm
20
.2
21
5
32
.3
17
.9
19
8
34
.0
12
.6
15
2
29
.5
6.
15
85
.7
27
.1
4 - 10
Third
Edition
Area Depth
Deslgnatlont A,mm2 d,mm
Flange
Web
Thick
·
ness
fw,
mm
Axis X·X Axis Y·Y
Thick
·
Width ness
b,,mm t,,mm
Ix Sx fx
106 mm" 103 mm3 mm
,, s,
,,
S610 x 180 22900
622
158 20100
622
149 19000
610
134 17100
610
119 15200
610
S510 x 143 18200
516
128 16400
516
112 14200
204
27.7
200
27.7
184
22.1
181
22.1
178
22.1
183
23.4
179
23.4
162
20.2
20.3
15.7
18.9
15.9
12.7
20.3
16.8
16.1
12.8
18.l
11.7
14.0
10.4
1320 4240
240
1230 3950
247
995 3260
229
938 3080
234
878 2880
240
700 2710
196
658 2550
200
530 2090
193
34.9 341
39.0
32.5 321
39.9
19.0 206
33.0
21.3 228
33.9
19.7 216
34.4
11.8 145
30.4
10.4 127

Deformations in a Transverse Cross Section
• Deformation due to bending moment M is
quantified by the curvature of the neutral surface
1 εm σm 1 Mc
= = =
ρ c
M
EI
Ec Ec I
=
• Although cross sectional planes remain planar
when subjected to bending moments, in-plane
deformations are nonzero,
= νy
ρ
= νy
ρ
ε = −νε ε = −νε
y x z x
• Expansion above the neutral surface and
contraction below it cause an in-plane curvature,
1 ν
= = anticlastic curvature
ρ
ρ′
4 - 11
Third
Edition

Sample Problem 4.2
SOLUTION:
• Based on the cross section geometry,
calculate the location of the section
centroid and moment of inertia.
I = ∑(I + Ad2
)
∑yA
Y = ′
x
∑ A
• Apply the elastic flexural formula to
find the maximum tensile and
compressive stresses.
Mc
σm =
I
A cast-iron machine part is acted upon
by a 3 kN-m couple. Knowing E =
GPa and neglecting the effects of
fillets, determine (a) the maximum
165 • Calculate the curvature
1 M
=
ρ EI
tensile and compressive stresses, (b)
the radius of curvature.
4 - 12
Third
Edition

Sample Problem 4.2
SOLUTION:
Based on the cross section geometry, calculate
the location of the section centroid and
moment of inertia.
3
∑yA = 114×10
Y = = 38 mm
∑ A 3000
= ∑(I + Ad 2
)=
∑(
1
bh3 + Ad 2
)
Ix′ 12
= 90×203 +1800×122 )+ ( 1 30×403 +1200
×182 )
(
1
12 12
I = 868×103mm = 868×10-9
m4
4 - 13
Third
Edition
Area, mm2 y, mm yA, mm3
1
2
20×90 =1800
40×30 =1200
50
20
90×103
24×103
∑ A = 3000 ∑ yA =114×103

Sample Problem 4.2
• Apply the elastic flexural formula to find the
maximum tensile and compressive stresses.
Mc
I
σ =
m
M cA 3 kN ⋅m×0.022m
σ = =
A
868×10−9 mm4
I
− M cB −3 kN ⋅m×0.038m
σ = =
B
868×10−9 mm4
I
• Calculate the curvature
1 M
EI
3 kN ⋅m
=
ρ
=
(165 GPa)(868×10-9 m4
)
4 - 14
Third
Edition
1 = 20.95×10−3m-1
ρ
ρ = 47.7 m
σB = −131.3 MPa
σ A = +76.0 MPa

Bending of Members Made of Several Materials
• Consider a composite beam formed from
two materials with E1 and E2.
Normal strain varies linearly.
•
y
ε = −
ρ
x
• Piecewise linear normal stress variation.
− E1y − E2 y
σ = E ε σ = E ε
= =
1 1 x 2 2 x
ρ ρ
Neutral axis does not pass through
section centroid of composite section.
Elemental forces on the section are
•
dF = σ dA = −
E1y
dA dF = σ dA = −
E2 y
dA
1 1 2 2
ρ ρ
My
σ x = − • Define a transformed section such that
I
( 1)
nE y
dA = −
E y E
1 (ndA) 2
σ1 = σ x σ2 = nσ x dF = − n =
2
ρ ρ E1
4 - 15
Third
Edition

Example 4.03
SOLUTION:
• Transform the bar to an equivalent cross
section made entirely of brass
• Evaluate the cross sectional properties of
the transformed section
• Calculate the maximum stress in the
transformed section. This is the correct
maximum stress for the brass pieces of
the bar.
Bar is made from bonded pieces of
steel (Es = 29x106 psi) and brass
(Eb = 15x106 psi). Determine the
maximum stress in the steel and
brass when a moment of 40 kip*in
is applied.
• Determine the maximum stress in the
steel portion of the bar by multiplying
the maximum stress for the transformed
section by the ratio of the moduli of
elasticity.
4 - 16
Third
Edition

Example 4.03
SOLUTION:
• Transform the bar to an equivalent cross section
made entirely of brass.
Es 29×10 psi
6
n = = =1.933
15×106 psi
Eb
bT = 0.4 in +1.933×0.75 in + 0.4 in = 2.25 in
Evaluate the transformed cross sectional properties
I = bT h3
= (2.25 in.)(3 in)3
•
1 1
12 12
= 5.063 in4
Calculate the maximum stresses
Mc (40 kip ⋅in)(1.5 in
)
•
σm = = =11.85 ksi
5.063 in4
I
(σb
)max
(σs
)
= σm
= nσm =1.933×11.85 ksi
4 - 17
Third
Edition
(σb )max =11.85
ksi
(σs )max = 22.9 ksi

Reinforced Concrete Beams
• Concrete beams subjected to bending moments are
reinforced by steel rods.
The steel rods carry the entire tensile load below
the neutral surface. The upper part of the
concrete beam carries the compressive load.
In the transformed section, the cross sectional area
of the steel, As, is replaced by the equivalent area
nAs where n = Es/Ec.
To determine the location of the neutral axis,
(bx)x − n A (d − x) = 0
•
•
•
s
2
1 b x2
+ n As x − n Asd = 0
2
• The normal stress in the concrete and steel
My
σ x = −
I
σc = σ x σs = nσ x
4 - 18
Third
Edition

Sample Problem 4.4
SOLUTION:
• Transform to a section made entirely
of concrete.
• Evaluate geometric properties of
transformed section.
• Calculate the maximum stresses
in the concrete and steel.
A concrete floor slab is reinforced with
5/8-in-diameter steel rods. The modulus
of elasticity is 29x106psi for steel and
3.6x106psi for concrete. With an applied
bending moment of 40 kip*in for 1-ft
width of the slab, determine the maximum
stress in the concrete and steel.
4 - 19
Third
Edition

Sample Problem 4.4
SOLUTION:
• Transform to a section made entirely of concrete.
Es 29×106 psi
n = = = 8.06
3.6×106 psi
Ec
nA = 8.06×2⎡π (5 in)2
= 4.95in
2
⎤
s ⎢
⎣ 4 8
• Evaluate the geometric properties of the
transformed section.
⎛
x ⎞
12x⎜ ⎟ − 4.95(4 − x) =
0
x =1.450in
⎝2
⎠
I = 1 (12in)(1.45in)3 + (4.95in2 )(2.55
in)2
= 44.4in4
3
Calculate the maximum stresses.
•
Mc1 40 kip ⋅in ×1.45
in
σ = =
c
44.4in4
I
n
Mc2 = 8.06
40 kip ⋅in ×2.55in
σ =
s
44.4in4
I
4 - 20
Third
Edition
σs = 18.52ksi
σc =1.306 ksi

Stress Concentrations
Mc
I
Stress concentrations may occur:
• in the vicinity of points where the
loads are applied
σ = K
m
• in the vicinity of abrupt changes
in cross section
4 - 21
Third
Edition

Plastic Deformations
• For any member subjected to pure bending
−
y
ε
ε strain varies linearly across the section
=
x m
c
• If the member is made of a linearly elastic material,
the neutral axis passes through the section centroid
My
σ x = −
and
I
• For a material with a nonlinear stress-strain curve,
the neutral axis location is found by satisfying
Fx = ∫σx dA = 0 M = ∫ − yσ x dA
• For a member with vertical and horizontal planes of
symmetry and a material with the same tensile and
compressive stress-strain relationship, the neutral
axis is located at the section centroid and the stress-
strain relationship may be used to map the strain
distribution from the stress distribution.
4 - 22
Third
Edition

Plastic Deformations
• When the maximum stress is equal to the ultimate
strength of the material, failure occurs and the
corresponding moment MU is referred to as the
ultimate bending moment.
• The modulus of rupture in bending, RB, is found
from an experimentally determined value of MU
and a fictitious linear stress distribution.
MU c
R =
B
I
• RB may be used to determine MU of any member
made of the same material and with the same
cross sectional shape but different dimensions.
4 - 23
Third
Edition

Members Made of an Elastoplastic Material
• Rectangular beam made of an elastoplastic material
Mc
σ x ≤ σY σm =
I
I
σm = σY MY = c σY = maximum elastic moment
• If the moment is increased beyond the maximum
elastic moment, plastic zones develop around an
elastic core.
y2
⎛ ⎞
3 1 Y
⎟
⎜
M = 2 MY ⎜1−
3
yY = elastic core half - thickness
⎟
c2
⎝ ⎠
• In the limit as the moment is increased further, the
elastic core thickness goes to zero, corresponding to a
fully plastic deformation.
M p = 2 MY = plastic moment
3
M p
k = = shape factor (depends only on cross section shape)
MY
4 - 24
Third
Edition

• Fully plastic deformation of a beam with only a
vertical plane of symmetry.
• The neutral axis cannot be assumed to pass
through the section centroid.
• Resultants R1 and R2 of the elementary
compressive and tensile forces form a couple.
R1 = R2
A1σY = A2σY
The neutral axis divides the section into equal
areas.
• The plastic moment for the member,
M p = AσY )d
(1
2
4 - 25
Third
Edition
Plastic Deformations of Members With a
Single Plane of Symmetry

Residual Stresses
• Plastic zones develop in a member made of an
elastoplastic material if the bending moment is
large enough.
• Since the linear relation between normal stress and
strain applies at all points during the unloading
phase, it may be handled by assuming the member
to be fully elastic.
• Residual stresses are obtained by applying the
principle of superposition to combine the stresses
due to loading with a moment M (elastoplastic
deformation) and unloading with a moment -M
(elastic deformation).
• The final value of stress at a point will not, in
general, be zero.
4 - 26
Third
Edition

Example 4.05, 4.06
A member of uniform rectangular cross section is
subjected to a bending moment M = 36.8 kN-m.
The member is made of an elastoplastic material
with a yield strength of 240 MPa and a modulus
of elasticity of 200 GPa.
Determine (a) the thickness of the elastic core, (b)
the radius of curvature of the neutral surface.
After the loading has been reduced back to zero,
determine (c) the distribution of residual stresses,
(d) radius of curvature.
4 - 27
Third
Edition

Example 4.05, 4.06
• Thickness of elastic core:
y2
⎛ ⎞
3 1 Y
⎟
Y
⎜
M = 2 M ⎜1−
3
2
⎟
c
⎝ ⎠
y2
⎛ ⎞
36.8kN⋅m = 3 (28.8kN⋅m)⎜1−1 Y
⎟
⎜ ⎟
2 3 2
c
⎝ ⎠
yY yY
= = 0.666 Y
c 60mm
• Radius of curvature:
• Maximum elastic moment: 240×106 Pa
σY
εY = = 9
E 200×10 Pa
= 2 bc2 = 2 (50×10−3m)(60×10−3
m)2
I
c =1.2×10−3
3 3
=120×10−6 m3
MY = σY = (120×10−6 m3 )(240
MPa)
yY
εY =
ρ
y
I
40×10−3m
c
= 28.8 kN⋅m
Y
ρ = =
1.2×10−3
εY
4 - 28
Third
Edition
ρ = 33.3m
2 y = 80 mm

Example 4.05, 4.06
• M = 36.8 kN-m
yY = 40 mm
• M = -36.8 kN-m
Mc 36.8kN ⋅m
• M = 0
At the edge of the elastic core,
σ′ = =
m
120×106 m3
I
σY −35.5×106 Pa
= 240MPa σx
εx = =
< 2σY
= 306.7 MPa 200×109 Pa
E
= −177.5×10−6
yY 40×10−3m
ρ = − =
177.5×10−6
εx
4 - 29
Third
Edition
ρ = 225m

Eccentric Axial Loading in a Plane of Symmetry
• Stress due to eccentric loading found by
superposing the uniform stress due to a centric
load and linear stress distribution due a pure
bending moment
(σ x )centric + (σ x
)bending
σ x =
P My
=
− A I
• Eccentric loading
F = P
M = Pd
• Validity requires stresses below proportional
limit, deformations have negligible effect on
geometry, and stresses not evaluated near points
of load application.
4 - 30
Third
Edition

Example 4.07
SOLUTION:
• Find the equivalent centric load and
bending moment
• Superpose the uniform stress due to
the centric load and the linear stress
due to the bending moment.
• Evaluate the maximum tensile and
compressive stresses at the inner
and outer edges, respectively, of the
superposed stress distribution.
An open-link chain is obtained by
bending low-carbon steel rods into the
shape shown. For 160 lb load, determine
(a) maximum tensile and compressive
stresses, (b) distance between section
centroid and neutral axis
• Find the neutral axis by determining
the location where the normal stress
is zero.
4 - 31
Third
Edition

Example 4.07
• Normal stress due to a
centric load
A = πc2
= π (0.25in
)2
= 0.1963in2
P 160lb
σ0 =
=
0.1963in2
A
= 815psi
• Equivalent centric load
and bending moment
P =160lb
M = Pd = (160lb)(0.6in)
=104lb⋅in
• Normal stress due to
bending moment
= 1 πc4 = 1 π
(0.25)4
I 4 4
−3
4
= 3.068×10 in
Mc (104lb⋅in)(0.25in
)
σ = =
m
.068×10−3 in4
I
= 8475psi
4 - 32
Third
Edition

Example 4.07
• Maximum tensile and compressive
stresses
• Neutral axis location
P My0
I
0 =
−
A
σt = σ0 +σm
= 815 +8475
= σ0 −σm
= 815−8475
3.068×10−3in4
P I
= (815
psi)
y0 =
σc A M 105lb ⋅in
4 - 33
Third
Edition
y0 = 0.0240in
σc = −7660 psi
σt = 9260 psi

Sample Problem 4.8
The largest allowable stresses for the cast
iron link are 30 MPa in tension and 120
MPa in compression. Determine the largest
force P which can be applied to the link.
SOLUTION:
• Determine an equivalent centric load and
bending moment.
• Superpose the stress due to a centric
load and the stress due to bending.
• Evaluate the critical loads for the allowable
tensile and compressive stresses.
From Sample Problem 2.4,
A = 3×10 m
Y = 0.038m
I = 868×10−9 m4
−3
2 • The largest allowable load is the smallest
of the two critical loads.
4 - 34
Third
Edition

Sample Problem 4.8
• Determine an equivalent centric and bending loads.
d = 0.038 − 0.010 = 0.028m
P =centric load
M = Pd = 0.028P =bending moment
• Superpose stresses due to centric and bending loads
P (0.028P
)(0.022)
P McA
σ A = −
+
= − + = +377 P
3×10−3 868×10−9
A
P
I
McA P (0.028P
)(0.022)
σB = −
−
= − − = −1559 P
3×10−3 868×10−9
A I
• Evaluate critical loads for allowable stresses.
σ A = +377 P = 30 MPa P = 79.6 kN
σB = −1559 P = −120 MPa P = 79.6
kN
• The largest allowable load
4 - 35
Third
Edition
P = 77.0 kN

Unsymmetric Bending
• Analysis of pure bending has been limited
to members subjected to bending couples
acting in a plane of symmetry.
• Members remain symmetric and bend in
the plane of symmetry.
• The neutral axis of the cross section
coincides with the axis of the couple
• Will now consider situations in which the
bending couples do not act in a plane of
symmetry.
• Cannot assume that the member will bend
in the plane of the couples.
• In general, the neutral axis of the section will
not coincide with the axis of the couple.
4 - 36
Third
Edition

Unsymmetric Bending
⎛−
y
σ
⎞d
A
• 0 = F = σ dA = ∫
⎜
m
⎟
∫
x x
c
⎝ ⎠
or 0 = y dA
∫
neutral axis passes through centroid
M = M = − y⎛− y σ
⎟dA m
⎞
• ⎜
∫
Wish to determine the conditions under
which the neutral axis of a cross section
of arbitrary shape coincides with the
axis of the couple as shown.
z
c
⎝ ⎠
σmI
or M = I = Iz = moment of inertia
c
defines stress distribution
• The resultant force and moment
from the distribution of
elementary forces in the section
must satisfy
Fx = 0 = My Mz = M = appliedcouple
0 = M = zσ dA = z⎜−
y
σ
⎛ ⎞
• ⎟d
A
∫ ∫
y x m
⎠
c
⎝
or 0 = yz dA =I yz = product of inertia
∫
couple vector must be directed along
a principal centroidal axis
4 - 37
Third
Edition

Unsymmetric Bending
Superposition is applied to determine stresses in
the most general case of unsymmetric bending.
• Resolve the couple vector into components along
the principle centroidal axes.
M z = M cosθ M y = M sinθ
• Superpose the component stress distributions
M y y
−
M z y
σ = +
x
I I
z y
• Along the neutral axis,
(M cosθ)y (M sinθ)y
− +
M y y
−
M z y
σ = 0 = + =
x
Iz I y Iz I y
y I z
tanφ =
=
tanθ
z I y
4 - 38
Third
Edition

Example 4.08
SOLUTION:
• Resolve the couple vector into
components along the principle
centroidal axes and calculate the
corresponding maximum stresses.
M z = M cosθ M y = M sinθ
• Combine the stresses from the
component stress distributions.
M y y
−
M z y
σ = +
x
I I
A 1600 lb-in couple is applied to a
rectangular wooden beam in a plane
forming an angle of 30 deg. with the
vertical. Determine (a) the maximum
stress in the beam, (b) the angle that the
neutral axis forms with the horizontal
plane.
z y
• Determine the angle of the neutral
axis.
y I z
tanφ =
=
tanθ
z I y
4 - 39
Third
Edition

Example 4.08
• Resolve the couple vector into components and calculate
the corresponding maximum stresses.
M z = (1600lb ⋅in)cos30 =1386lb ⋅in
M y = (1600lb ⋅in)sin 30 = 800lb ⋅in
Iz = (1.5in)(3.5in)3
= 5.359in4
1
12
I y = (3.5in)(1.5in)3
= 0.9844
in4
1
12
The largest tensile stress due to M z occurs along AB
M z y (1386lb ⋅in)(1.75
in)
σ = = = 452.6psi
1
5.359in4
Iz
The largest tensile stress due to M z occurs along AD
(800lb⋅in)(0.75in
)
M y z
σ = = = 609.5psi
2
0.9844in4
I y
• The largest tensile stress due to the combined loading
occurs at A.
σmax = σ1 +σ2 = 452.6 +
609.5
4 - 40
Third
Edition
σmax =1062 psi

Example 4.08
• Determine the angle of the neutral axis.
4
Iz 5.359in
tan 30
tan
φ
tanθ
= =
0.9844in4
I y
= 3.143
4 - 41
Third
Edition
φ =
72.4o

General Case of Eccentric Axial Loading
• Consider a straight member subject to equal
and opposite eccentric forces.
• The eccentric force is equivalent to the system
of a centric force and two couples.
P = centric force
M y = Pa M z = Pb
• By the principle of superposition, the
combined stress distribution is
M y z
P M z y
σ =
−
A
+
x
I I
z y
• If the neutral axis lies on the section, it may
be found from
M y
M z P
A
y − z =
I z I y
4 - 42
Third
Edition

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