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Lecture notes, lecture 17 - Bending
Mechanics of Material (University of Sheffield)
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Chapter 12
Bending
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Objectives
• To determine the stress in elastic symmetric
members subjected to bending
• To develop method to determine the stress in
unsymmetric beams subjected to bending
• To determine the stress in composite beams
subjected to bending
Chapter 12 – Bending
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Bending Deformation
In this
hi section,
i we will
ill discuss
di
the
h
deformations that occur when a straight
prismatic beam, made of a
homogeneous materials
materials, is subjected to
bending.
The discussion starts for beams having a
cross-sectional area that is symmetrical
with respect to an axis, and the bending
moment is applied about an axis
perpendicular
di l to
t this
thi axis
i off symmetry
t
as show in the figure.
The behavior of members having
unsymmetrical cross-sections of made
from several different materials will
also be discussed.
Chapter 12 – Bending
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Bending Deformation
Figure (a) shows undeformed bar having a
square cross section and marked with
longitudinal and transverse grid lines.
When a bending moment is applied, it
tends to distort these lines into the
pattern shown in figure (b). The
longitudinal lines become curved and
the vertical transverse lines remains
straight and yet undergo a rotation.
The bending moment causes the material
within the bottom portion of the bar to
stretch and the material within the top
portion to compress → there will be a
surface in which longitudinal fibers of
the material will not undergo a change
in length. These surface is called
neutral surface.
Chapter 12 – Bending
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Bending Deformation
From the observations, the following
assumptions regarding the way the stress
deform the material are made:
·Longitudinal axis x, which lies within the
neutral axis will not experience any change of
length. It will only deform into a curve that lies
i the
in
h x-y plane
l
off symmetry.
·All cross-section of the beam remain plane
and perpendicular to the longitudinal axis
during the deformation
deformation.
·Any deflection of the cross section within its
own plane will be neglected. The z-axis, lying
in the plane of the cross section and about
which the cross section rotation is called the
neutral axis.
Chapter 12 – Bending
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Strain due to Bending
A segment of the beam
beam, which is located a
distance x along the beam’s length and has a
undeformed thickness ∆x is isolated.
Any li
A
line segment ∆x,
∆ located
l
d on the
h neutrall
surface, does not change its length, whereas
any line segment ∆s, located at the arbitrary
distance y above the neutral surface
surface, will
contract, and becomes ∆s’ after deformation.
Before deformation, ∆s= ∆x. After deformation
∆x has a radius of curvature ρ, with a center
of curvature at point O’.
Since ∆θ defines the angle
g between the crosssectional sides of the element, ∆x= ∆s=ρ∆θ.
The deformed length of ∆s becomes ∆s’=(ρy) ∆θ. The normal strain along ∆s:
∆s '−∆s
( ρ − y )∆θ − ρ∆θ
= lim
∆s →0
∆θ →0
ρ∆θ
∆s
ε = lim
Chapter 12 – Bending
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or
ε =−
y
p
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Strain due to Bending
For any specific cross section,
section the
longitudinal normal strain will vary
linearly with y from the neutral axis. A
contraction ((-ε)) will occur in fibers
located above the neutral axis (+y),
whereas elongation (+ε) will occur in
fibers located below the axis (-y).
The maximum strain occurs at the
outermost fiber, located at the distance c
from the neutral axis.
⎛ y⎞
⎝c⎠
ε = −⎜ ⎟ε max
This strain is in the longitudinal
g
direction
(ε x ). Due to the Poisson’s effect, there
should also be strains in the lateral
directions ( ε y = −vε x , ε z = −vε x ), which
deforms the plane of cross sectional
area.
Chapter 12 – Bending
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Strain due to Bending
If the material behaves in a linearelastic manner → Hooke
Hooke’ss law (σ=Eε)
is valid → A linear variation of normal
strain will cause a linear variation of
normal stress.
y
c
σ = −( )σ max
The
equation
represents
stress
distribution over the cross sectional
area. For positive M, compressive stress
(-σ) occurs above the neutral axis (+y)
and
d tensile
il stress (+σ)
( ) occurs below
b l the
h
neutral axis (-y).
Chapter 12 – Bending
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Neutral Axis
The neutral axis on the cross section can
be located by satisfying the condition that
the resultant force produced by the stress
distribution over the cross
cross-sectional
sectional area
must be equal to zero. Noting that dF=σdA
acts on arbitrary element dA, we require
FR = ∑ Fx ;
∫
⎛ y⎞
=
=
−
dF
σ
dA
A
∫A
∫ A ⎜⎝ c ⎟⎠σ max dA
=−
σ max
c
∫
A
ydA
dA = 0
Since σ max / c is not equal to zero, then
∫
A
ydA = 0
This condition can only be satisfied if the neutral axis is also the horizontal
centroidal axis for the cross section.
Chapter 12 – Bending
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The Flexure Formula
The stress
Th
t
in
i the
th beam
b
can be
b determined
d t
i d
from the requirement that the resultant
internal moment M must be equal to the
moment produced by the stress distribution
about the neutral axis. The moment of dF
about the neutral axis is dM=-ydF.
(M R )z = ∑ M z ;
∫
⎛ y
⎞
−
=
−
=
−
−
ydF
y
(
σ
dA
)
y
σ
⎜
A
∫A
∫ A ⎝ c max ⎟⎠dA = M
⇒M =
Since
σ max
∫
c
∫
A
y 2 dA = 0
2
A y dA = I , then σ max =
Mc
I
Substituting
Myy
⎛ y⎞
σ max → σ = −
I
⎝c⎠
σ = −⎜
Chapter 12 – Bending
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Example 12.2
The simple supported beam has
the cross-sectional
cross sectional area as
shown. Determine the absolute
maximum bending stress in the
beam and draw the stress
distribution over the crosssection at this location.
Chapter 12 – Bending
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Example 12.2
Maximum Internal Moment
The maximum internal bending moment in
the beam, M=22.5kNm, occurs at the
center
t as shown
h
in
i the
th bending
b di momentt
diagram.
Section Property
Because of symmetry, the centroid C and
thus the neutral axis pass through the
midheight of the beam.
I = ∑ (T + Ad 2 )
⎡1
⎤
= 2 ⎢ (250)(20)3 + (250)(20)(160) 2 ⎥ +
⎣12
⎦
1
(20)(300) 3
12
= 301.3 × 106 mm 4
Chapter 12 – Bending
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Example 12.2
Bending Stress
The absolute maximum bending stress
is
Mc
σ max =
σ max
I
22.5 ×106 (170)
=
= 12.7 MPa
301.3 ×106
Two- and three- dimensional view of
the stress distribution are shown.
My B
22.5 ×106 (150)
;σ B = −
σB = −
= −11.2 MPa
I
301.3 ×106
At point B,
The normal stress acting on elements of material located at point B and D is
shown.
Chapter 12 – Bending
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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 p
plane of symmetry.
y
y
• The neutral axis of the cross section
coincides with the axis of the couple.
Chapter 12 – Bending
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Unsymmetric Bending
• Consider situations in which the
bending couples do not act in a
plane of symmetry.
• Th
The member
b will
ill nott bend
b d in
i the
th
plane of the couples.
• In general, the neutral axis of the
section will not coincide with the
axis of the couple.
Chapter 12 – Bending
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Unsymmetric Bending
Superposition is applied to determine
stresses in the most general case of
y
bending.
g
unsymmetric
·Resolve the couple vector into
components along the principle
centroidal axes.
M z = M cos θ ; M y = M sin θ
·Superpose the component stress
distributions
Mz y M yz
σx = −
+
Iy
Iz
Chapter 12 – Bending
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Unsymmetric Bending
Orientation of neutral axis
There is
Th
i no normall stress
t
att neutral
t l
axis, therefore
Mz y M yz
σx = 0 = −
+
Iz
Iy
=−
( M cos θ ) y ( M sin θ ) z
+
Iz
Iy
tan α =
y Iz
= tan θ
z Iy
Chapter 12 – Bending
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Example 12.5
The rectangular cross section is subjected to a bending moment
of M=12kNm. Determine the normal stress developed at each
corner
co
e of
o thee section,
sect o , and
a d specify
spec y thee orientation
o e at o of
o thee neutral
eut a
axis.
Chapter 12 – Bending
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Example 12.5
Internal Moment Components
The moment is resolved into its y and z
components, where
4
M y = − (12kNm) = −9.6kNm
5
3
M z = (12kNm) = 7.2kNm
5
Section Properties
The moment of inertia about the y and z axes are
1
(400)(200) 3 = 266.7 ×10 6 mm 4
12
1
I z = (200)(400) 3 = 1067 × 106 mm 4
12
Iy =
Chapter 12 – Bending
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Example 12.5
Bending Stress
Mz y M yz
σ =−
+
Iz
Iy
7.2 × 10 6 (200) − 9.6 × 106 (−100)
σB = −
+
= 2.25MPa
1067 × 10 6
266.7 × 106
7.2 × 10 6 (200) − 9.6 × 106 (100)
+
= −4.95MPa
σC = −
1067 × 106
266.7 × 106
7.2 × 10 6 (−200) − 9.6 ×10 6 (100)
= −2.25MPa
σD = −
+
1067 × 10 6
266.7 ×10 6
7.2 ×10 6 (−200) − 9.6 × 106 (−100)
+
σE = −
= 4.95MPa
1067 ×10 6
266.7 × 106
Orientation of Neutral Axis
The location z of the neutral axis
can be established by proportion
Chapter 12 – Bending
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2.25MPa
4.95MPa
=
z
(200mm − z )
z = 62.5mm
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Example 12.5
The angle α is used to specify the orientation
of neutral axis (N.A.) with respect to z or
maximum principal axis.
axis
From sign convention, θ must be measured
from the +z axis toward the +y axis.
θ = − tan −1
4
= −53.1o
3
Thus,
tan α =
Iz
tan θ
Iy
1067 ×106
tan α =
tan(−53.1o )
6
266.7 ×10
α = −79.4o
Chapter 12 – Bending
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Composite Beams
Beams constructed
B
t t d off two
t
or more
materials are referred to as composite
beams. Composite beams are used to
develop a more efficient means for
carrying applied loads.
The flexure formula derived previously
was developed
d l d for
f beams
b
whose
h
material
i l
is homogeneous → not applicable for
composite beams.
Transformed section method is used to
modify or transform the cross section of
composite beam into one made of single
material. Once it is done, the flexure
formula derived previously can be applied
for the stress analysis.
Chapter 12 – Bending
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Composite Beams
• Consider a composite beam formed
from two materials with E1 and E2
with E1 > E2 .
• Normal strain varies linearly: ε x = −
y
p
• Piecewise linear normal stress variation:
σ 1 = E1ε x = −
E1 y
ρ
, σ 2 = E 2ε x = −
E2 y
ρ
Notice the “jump” in stress at the juncture
of the materials.
Neutral
N
t l axis
i does
d
nott pass through
th
h section
ti
centroid of composite section.
Chapter 12 – Bending
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Transformed-Section Method
·Elemental forces on the section are:
dF1 = σ 1dA = −
E1 y
dF2 = σ 2 dA = −
E2 y
ρ
ρ
dA
dA
· Define a transformed section such that
dF1 = −
n=
(nE2 ) y
ρ
dA = −
E2 y
ρ
(ndA
dA)
E1
E2
·Stresses
S
iin the
h two materials
i l
My
I
σ 2 = σ x , σ 1 = nσ x
σx = −
Chapter 12 – Bending
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Example
A composite beam is made of wood and reinforced with a steel
pplate located on the bottom of side. It has the cross-sectional area
as shown. If the beam is subjected to a bending moment of M = 2
kNm, determine the normal stress at points B and C. Take Ew = 12
GPa and Est = 200 GPa
Chapter 12 – Bending
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Example
Section Properties
Although the choice is arbitrary, here we
transform the section into one made
entirely of steel.
Since steel has a greater stiffness than
wood ( Est > Ew ),
) the width of wood will
be reduced to an equivalent width of steel.
n=
Ew
12
(150mm) = 9mm
→ bst = nbw =
Est
200
The location of centroid (bottom of section is taken as a reference)
y=
∑ ( y A) = 10(20 ×150) + 95(9 ×150) = 36.38mm
20 ×150 + 9 ×150
∑A
The moment of inertia (about the neutral axis)
⎡1
⎤ ⎡1
⎤
I = ⎢ 150(20) 3 + (150 × 20)(36.38 − 10) 2 ⎥ + ⎢ 9(150) 3 + (150 × 9)(95 − 36.38) 2 ⎥
⎣12
⎦ ⎣12
⎦
= 9.358 ×106 mm 4
Chapter 12 – Bending
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Example
Normal stress
Normal stress at B’ and C is
My B '
M
2000(170 − 36.38)
=−
= −28.6 MPa
I
9.358 × 106
My
2000(−36.38)
σC = − C = −
= 7.78MPa
I
9.358 × 106
σ B' = −
Normal stress distribution on the
transformed (all steel) section is shown.
The normal stress in the wood, located at B,
is
σ B = nσ B ' =
12
(−28.6) = −1.71MPa
200
Normal stress distribution in the actual
beam is shown.
Chapter 12 – Bending
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Reinforced Concrete Beams
Concrete beams subjected to bending moments are
reinforced by steel rods.
The steel rods carry
y the entire tensile load below the
neutral surface. The upper part of the concrete beam
carries the compressive load.
In the transformed section
section, the cross sectional area
of the steel, As, is replaced by the equivalent area nAs
where n = Es .
Ec
To determine
d
i the
h location
l
i off the
h neutrall axis,
i
(bh' )
h'
− nAst (d − h' ) = 0
2
1 2
bh' + nAst h'− nAst d = 0
2
The normal stress in the concrete and steel
My
I
σ c = σ x , σ s = nσ x
σx = −
Chapter 12 – Bending
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Example
The reinforced concrete beam has the cross-sectional
area as shown.
h
If it is
i subjected
bj t d to
t a bending
b di momentt
of M=60kNm, determine the normal stress in each steel
reinforcing rods and the maximum normal stress in the
concrete. Take Est = 200GPa and Econc = 25GPa
Chapter 12 – Bending
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Example
Since beam is made from concrete, its strength
in supporting tensile stress is neglected.
Section Properties
p
The steel will be transformed into an equivalent
area of concrete.
n=
Est
Econc
A' = nAst =
[
]
200
2 × π (12.5) 2 = 7856mm 2
25
The location of centroid
h' ⎞
⎛h
− 7856(400h' ) = 0
300h' ⎜
2
⎝ ⎠
h'2 +52.37 h'−20949.33 = 0
h' = 120.9mm
Chapter 12 – Bending
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Example
The moment of inertia (about the neutral axis)
2
⎡1
⎛ 120.9 ⎞ ⎤
3
2
I = ⎢ 300(120.9) + (300 × 120.9)⎜
⎟ ⎥ + 7856(400 − 120.9)
⎝ 2 ⎠ ⎦⎥
⎣⎢12
[
]
= 788.67 × 106 mm 4
Normal stress
The maximum normal stress in the concrete is
(σ conc )max
Mc
60 × 106 (120.9)
=−
=−
= −9.20 MPa
I
788.67 ×106
The normal stress resisted by the “concrete”
strip, which replaced the steel, is
60 ×10 6 [− (400 − 120.9)]
My
=−
= 21.23MPa
σ 'conc = −
788.67 ×10 6
I
The normall stress
Th
t
i each
in
h off the
th two
t
reinforcing rods is therefore
σ stt = nσ 'conc =
200
(21.23) = 169.84 MPa
25
Chapter 12 – Bending
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