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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
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
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
Symmetrical and anti-symmetrical cross sections
• Are these also principal axes?
Equilateral triangle Open channel section
Z- sectionAngle section
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
Rectangular cross section• Maximum bending stress
2max||6
bhM X=σ
Cantilever beam with rectangular cross section
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
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
−=⇒−=
−==
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
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.
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.
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
−=
−
− − −= =
−
−= = −
−
.
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
Total displacement
αδ
α
cos
tan
22 vvu
vu
=+=
−=
Components of deflection of a nonsymmetrically loaded beam.
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
==
===
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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