Bending of straight beams - UFL · PDF fileBending of straight beams ... concentrated load P=...
Transcript of Bending of straight beams - UFL · PDF fileBending of straight beams ... concentrated load P=...
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Bending of straight beams• In mechanics of materials we cover symmetrical
cross 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 no
twist. 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
yMI
xM+−=σ
0=zzσ
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Rectangular cross section• Maximum bending stress
2max||6
bhM X=σ
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
σ
σ
σ0
cybxaE
ycxba
zz
zzzz
zz
++=∈=
′+′+′=∈
σσ
yIII
IMIMx
IIIIMIM
xyyx
xyyyx
xyyx
xyxxyzz ⎟
⎟⎠
⎞⎜⎜⎝
⎛
−
++⎟
⎟⎠
⎞⎜⎜⎝
⎛
−
+−= 22σ
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Moments
• Moment is perpendicular to the plane of the loads. If the plane of the loads makes an angle φwith the x-axis,
φφ
φφ
cotcot
cossin
xyx
y
yx
MMMM
MMandMM
−=⇒−=
−==
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Neutral axis
• For bending stress to be zero
φφ
α
α
ασ
cotcot
tan
tan
tan
xyy
xxy
xyyyx
xyxyx
xyyyx
xyxyxzz
IIII
IMIMIMIM
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 kNlies in the plane with an angle φ = π/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
ImmImmApropertiesSection
y
xyx
==
===
rad
II
IIII
AxisNeutralLocate
y
x
xyy
xxy
6407.07457.0tan
5774.0
3sin
3cos
3cot
3
cotcotcot
tan
:
−=⇒−=
=
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛
=⎟⎟⎠
⎞⎜⎜⎝
⎛⇒=
−=−−
=
αα
π
πππφ
φφφ
α
mkNMMmkNPM
Moments
x .18.31sin.0.3600.3
:
−==−=−=
φ
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 tan133.7 MPa
tan
Stress at B(70,82)
M 82 70 tan105.4 MPa
tan
zzx xy
Ax xy
Bx xy
y xI I
I I
I I
ασ
α
ασ
α
ασ
α
−=
−
− − −= =
−
−= = −
−
.
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Deflections
• Determine separately x and ycomponents of displacement. Here we show y component
• Curvature
)()tan(
11
22
2
2
2
xyyx
xyyyx
xyx
x
yy
zz
y
IIIEIMIM
IIEM
dzvd
dzvd
Rwhere
RR
−
+−=
−−=
≈∈
=−
α
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Total displacement
αδ
α
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 φ = 5π/9 with the x-axis. Locate points of maximum tensile and compressive stresses and magnitude of stresses. Find the maximum deflection. E=72 GPa.
460
462
10x73.300.82
010x69.3610000:
mmImmy
ImmImmApropertiesSection
y
xyx
==
===
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Solution
MPaII
M
MPaII
M
IIxyM
xyx
xB
xyx
xA
xyx
xzz
2.87tan
)tan)70(82(
8.63tan
)tan)70(118(
tan)tan(
:Stress
−=−−
=
=−
−−−=
−−
=
αασ
αασ
αασLocate Neutral Axis:
tan cot 5 9
tan 0.2277 0.2239
= − ⇒ =
= ⇒ =
x
y
II
rad
α φ φ π
α α
mmvvu
mmvu
mmEI
PLv
EI
x
95.6cos
54.1tan
78.648
sin48PL
:beam supportedSimplyDeflection
22
3
3
==+=
−=−=
==
=∆
αδ
α
φ
mkNMM
mkNPLM
x .85.25sin
.25.264
:Moment
==
==
φ
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Reading assignmentSections 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