Download - Straight Bending Beam
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Bending of straight beams
In mechanics of materials we cover symmetricalcross sections and bending in one plane. Now
we will consider the more general case
Moment perpendicular to a plane at an angle phi
from x-z plane (plane of loads). Centroidal axes.
Cantilever beam with an arbitrary cross section subjected to pure bending
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Shear loading
We assume for now pure bending with notwist. This implies shear forces passing
through shear center
Cantilever beam with an arbitrary cross section subjected to shear loading
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Symmetrical bending
Moments of inertia
Moments of inertia also a tensor so has
principal axes
AdxyI
AdrJ
AdxI
AdyI
xy
y
x
==
=
=
2
2
2
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Symmetrical and anti-symmetrical
cross sections
Are these also principal axes?
Equilateral triangle Open channel section
Z- sectionAngle section
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Symmetrical bending
Euler-Bernoulli beam theory, Leonhard Euler(1707-1783) and Daniel Bernoulli (1700-1782)
What are the assumptions?
For symmetrical cross section
Neutral axis
X
X
Y
Yzz
I
yM
I
xM+=
0=zz
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Rectangular cross section
Maximum bending stress
2max
||6
bh
MX
=
Cantilever beam with rectangular cross section
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Unsymmetrical bending
Equations of equilibrium
Plane sections remain plane
Combining it all
=
=
=
dAxM
dAyM
dA
zzy
zzx
zz
0
cybxaE
ycxba
zz
zzzz
zz
++=
=
++=
yIII
IMIMxIII
IMIMxyyx
xyyyx
xyyx
xyxxyzz
++
+= 22
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Moments
Moment is perpendicular to the plane of the
loads. If the plane of the loads makes an angle with the x-axis,
cotcot
cossin
xy
x
y
yx
MMM
M
MMandMM
==
==
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Neutral axis
For bending stress to be zero
cot
cottan
tan
tan
xyy
xxy
xyyyx
xyxyx
xyyyx
xyxyx
zz
II
II
IMIM
IMIM
xxIMIMIMIM
=
+
+=
=
++=
Pure bending of a nonsymmetrically loaded
cantilever beam
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Example 7.3 A cantilever beam of length 3m as shown in the figure
has a channel section. A concentrated load P=12.0 kN
lies in the plane with an angle =/3 with the x-axis.
The plane of the loads passes through the shear center
C. Locate points of maximum tensile and compressive
stresses and find the magnitude of stresses.
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Location of max stresses
460
462
10x73.300.82
010x69.3610000
:
mmImmy
ImmImmA
propertiesSection
y
xyx
==
===
rad
I
I
II
II
AxisNeutralLocate
y
x
xyy
xxy
6407.07457.0tan
5774.0
3
sin
3cos
3cot
3
cotcot
cottan
:
==
=
=
=
=
=
mkNMM
mkNPM
Moments
x .18.31sin
.0.3600.3
:
==
==
Since moment is negative, the part
above N.A is in tension and the bottom
part is in compression. Therefore
maximum tensile stress occurs at
point A and maximum compressive
stress at point B.
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Maximum stresses
( )
( )( )
( )( )
x
x
x
Stress at A(-70,-118)
M tan
tan
M 118 70 tan133.7MPa
tan
Stress at B(70,82)M 82 70 tan
105.4MPatan
zz
x xy
Ax xy
B
x xy
y x
I I
I I
I I
=
= =
= =
.
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Deflections
Determine separatelyxand y
components of displacement.
Here we show ycomponent
Curvature
)()tan(
11
22
2
2
2
xyyx
xyyyx
xyx
x
yy
zz
y
IIIE
IMIM
IIE
M
dz
vd
dz
vd
Rwhere
RR
+=
=
=
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Total displacement
cos
tan
22 vvu
vu
=+=
=
Components of deflection of a
nonsymmetrically loaded beam.
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Example 7.6
A simply supported beam of length 3m has a channel section. A
concentrated load P= 35.0 kN applied at the center of the beam, lies in
a plane with an angle = 5/9 with the x-axis. Locate points ofmaximum tensile and compressive stresses and magnitude of
stresses. Find the maximum deflection. E=72 GPa.
46
0
462
10x73.300.82
010x69.3610000
:
mmImmy
ImmImmA
propertiesSection
y
xyx
==
===
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Solution
MPaII
M
MPaII
M
II
xyM
xyx
xB
xyx
xA
xyx
xzz
2.87tan
)tan)70(82(
8.63tan
)tan)70(118(
tan
)tan(:Stress
=
=
=
=
=
Locate Neutral Axis:
tan cot 5 9
tan 0.2277 0.2239
= =
= =
x
y
I
I
rad
mmvvu
mmvu
mmEI
PLv
EI
x
95.6cos
54.1tan
78.648
sin48
PL
:beamsupportedSimplyDeflection
22
3
3
==+=
==
==
=
mkNMM
mkNPL
M
x .85.25sin
.25.264
:Moment
==
==
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Reading assignment
Sections 7.3-5: Question: Consider a horizontal cantilever
beam under a tip vertical load. What condition is required so
that tip will move both down and sideways?
Source: www.library.veryhelpful.co.uk/ Page11.htm