REVIEW 1- Bending Open and Closed Sectio
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Transcript of REVIEW 1- Bending Open and Closed Sectio
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7/25/2019 REVIEW 1- Bending Open and Closed Sectio
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AERSP 301Bending of open and closed section beams
Dr. Jose Palacios
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Direct stress at a point in the csdepends on!
" #ts location in the cs
" $he loading
" $he geometr% of the cs
Ass&mption "plane sectionsremain plane after deformation '(o)arping*+ or cross,section does notdeform in plane 'i.e. -+ -%% / 0*
Sign onentions2
egson pp 451
Direct stress calc&lation d&e to bending
" bending moment
S " shear forceP " aial load
$ " tor6&e
) " distrib&ted load
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Direct stress calculation due to bending (contd)
Beam s&b7ect to bending moments and %and bends abo&t its
ne&tral ais '(.A.*
(.A. " stresses are 8ero at (.A.
" centroid of cs 'origin of aes ass&med to be at *.
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(e&tral S&rface Definition
#n the process of bending there is an aial
line that do not etend or contract. $he
s&rface described b% the set of lines that
do not etend or contract is called thene&tral s&rface. 9ines on one side of the
ne&tral s&rface etend and on the other
contract since the arc length is smaller
on one side and larger on the other side
of the ne&tral s&rface. $he fig&re sho:s
the ne&tral s&rface in both the initial and
the bent config&ration.
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$he aial strain in a line element a distance yaboe the ne&tral s&rfaceis gien b%!
onsider element A at a distance ; from the (.A.
Direct Stress!
Beca&se < 'bending radi&s of c&rat&re* relates the strain to the
distance to the ne&tral s&rface!
Direct stress calculation due to bending (contd)
( )
=
=
=
0
0
l
llz
E
E zzz ==
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First Moment of Inertia Definition
=ien an area of an% shape+ and diision of that area into er% small+
e6&al,si8ed+ elemental areas 'dA*
and gien an ,%ais+ from :here each elemental area is located 'yi
andxi*
$he first moment of area in the >?> and >@> directions are respectiel%!
dAxAxI
ydAAyI
y
x
====
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# the beam is in p&re bending+ aial load res<ant on the cs is 8ero!
1stmoment of inertia of the cs abo&t the (.A. is 8ero N.A.passes through the centroid, C
Ass&me the inclination of the (.A. to xis
Direct stress calculation due to bending (contd)
$hen
$he direct stress becomes!
== AA z dAdA 00
cossin yx +=
( )
cossin yxEE
z +==
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Direct stress calculation due to bending (contd)
oment Res<ants!
S&bstit&ting for zin the aboe epressions for Mxand My+ and &sing
definitions for Ixx, Iyy, Ixy
dAxM
ydAM
zy
zx
=
=
dAxyI
dAxI
dAyI
xy
yy
xx
=
=
=
2
2
=
+=
+=
cos
sin
cossin
cossin
xyyy
xxxy
y
x
xyyyy
xxxyx
II
IIE
M
M
IE
IE
M
IE
IE
M
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Direct stress calculation due to bending (contd)
Csing the aboe e6&ation in!
=ies!
rom atri
orm
=
y
x
xyyy
xxxy
M
M
II
IIE1
cos
sin
=
y
x
xyyy
xxxy
xyyyxx M
M
II
II
III
E2
1
cos
sin
( )
cossin yxE
z
+=
y
III
IMIMx
III
IMIM
xyyyxx
xyyyyx
xyyyxx
xyxxxy
z
+
=
22
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Direct stress calculation due to bending (contd)
r+ rearranging terms!
#f My= 0+ Mxprod&ces a stress that aries :ith both and %. Similarl%for My+ if Mx/0.
#f the beam cs has either Cxor Cy'or both* as an ais of s%mmetr%+then Ixy/ 0.
$hen!
( ) ( )22
xyyyxx
xyxxy
xyyyxx
xyyyx
zIII
yIxIM
III
xIyIM
+
=
yy
y
xx
xz
I
xM
I
yM+=
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&rther+ if either %or is 8ero+ then!
)e sa: that the (.A. passes thro&gh the centroid of the cs. B&t
:hat abo&t its orientation
At an% point on the (.A. -8/ 0
Direct stress calculation due to bending (contd)
xx
x
zI
yM=
yy
y
zI
xM=or
tan
022
=
=
=
+
=
xyyyyx
xyxxxy
xyyyxx
xyyyyx
xyyyxx
xyxxxy
z
IMIM
IMIM
x
y
yIII
IMIMx
III
IMIM
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Eample Problem
$he beam sho:n is s&b7ected to a 1F00
(m bending moment in the ertical
plane.
alc&late the magnit&de and location ofma -8.
! mm
" mm ! mm
! mm
! mm
#st$ Calculate location of Centroid
mm528808120
)880(40)8120(60=
++==
xx
xxxx
A
Axxc
mm4.66
8808120
)880(40)8120(84=
+
+==
xx
xxxx
A
Ayyc
%
&
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Eample Problem 'contGd* alc&late #..+ #%%+ #.%+ :ith respect to .%!
462
3
23
23
mm1009.1)404.66(80812808
)4.6684(812012
8120
12
=+
++
=+= cxx dAbtI
4623
23
23
mm1031.1)4052(80812
808
)5260(8120
12
8120
12
=+
++
=+= cyy dAtbI
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Eample Problem 'contGd*
46 mm1034.0)4.6640()5240(808
)4.6684()5260(8120
=
+== A
xy xydAI
M% ' # Nm, M&'
( ) ( )22
xyyyxx
xyxxy
xyyyxx
xyyyx
z
III
yIxIM
III
xIyIM
+
=
mm]inyx,N/mmin[
39.05.1
2
z
xyz = & inspection, MA* at& ' +." mm and % ' +! mm(Ma% stress al-a&s further a-a&
From centroid)
%
&
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Deflections due to bendingDeflections due to bending
rom strength of materials+ recall that Hegson h. 15.I.F!
Beam bends abo&t its (.A. &ndermoments + %.
" Deflection normal to (.A. is Kentroid
moes from #'initial* to A'final*.
" )ith R as the center of c&rat&re and
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Deflections d&e to bending 'contGd*
&rther+
Beca&se
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Deflections d&e to bending 'contGd*
#nerse relation!
learl% prod&ces c&rat&res 'deflections* in 8 and %8 planeseen :hen % / 0 'and ice,ersa*
So an &ns%mmetrical beam :ill deflect erticall% and hori8ontall%
een :hen loading is entirel% in ertical 'or hori8ontal* plane.
/hat if I ha0e something s&mmetric11 i2e NACA #3 airfoil1
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Deflections d&e to bending 'contGd*
#f or % 'or both* are aes of s%mmetr% then #% / 0. $hen the
epressions simplif% to!
Starting :ith the general epression!
and integrating t:ice %o& can calc&late the disp. & in the ,direction
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Deflections d&e to bending 'contGd*
onsider the case :here a
do:n:ard ertical force+ )+ is
applied to the tip of a beam.
)hat is the tip deflection of the
beam
#ntegrating+
#ntegrating again
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Deflections d&e to bending 'contGd*
Csing b.c.Gs! L 8/0& / 0+ &G / 0
" =ies! A / B / 0
$h&s+
" $ip deflection!
#f the cs has an ais of s%mmetr%+ #%/ 0
4ou should do this on &our o-n
(z = L)
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Simplifications for thin,:alled sections
$hin,:alledt MM cs dimensions.
" Stresses constant thro&gh thicNness
" $erms in tI+ t3+ etcO neglected
#n that case #red&ces to!
)hat abo&t #%for this cs
)hat abo&t #%%for this cs
hori8ontal members
ertical members
5%ample$
4ou should do this on &our o-n
D bl 6 t i l C 6 ti
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Doubl& 6&mmetrical Cross+6ection
eam ending
eam has a fle%ural rigidit&$ 5I
&p(z)
7
pLA
LvEI
=
= 0)(
cs
2
0)(
2pLB
LvEI
=
=
0
0)0(
=
=
C
vEI
0
0)0(
=
=
D
EIv
D bl 6 t i l C 6 ti
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Doubl& 6&mmetrical Cross+6ection
eam ending
eam has a fle%ural rigidit&$ 5I
&
7
FA
FLvEI
=
= )(
cs
FLB
LvEI
== 0)(
0
0)0(
=
=
C
vEI
0
0)0(
=
=
D
EIv
F
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89:I8NA$ Macaule&s Method ;ead Megson Chp #.)
y
z
a a a a
W W 2W
A
C D F
Ra Rf
Determine the position and magnitude
of the ma%imum up-ard and do-n-ard
deflection of the beam$
(upa!")
4
3WRA= ("ona!")
4
3WRF=
]3[2]2[][ azWazWazWzRM A ++=
:he bending moments around the left hand side at an&
section < bet-een D and Fis$
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Concentrated and 9artial 6pan oads
Diract delta f&nction!
Eampleertical force of magnit&de F0locater at 9I
eaiside step f&nction!
Eampleertical distrib&ted force of magnit&de f0(z)oer the
second part of the beam onl%
)/()()( Lzzstpyfzf oy =
)2/( LzF = 7
f
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Concentrated and 9artial 6pan oads 5%ample
7
&
=>=> =>
f
Fo
7
%
=>=> =>
f
Mo
+
=
3
2
33)(####
Lzstp
Lzstpf
LzFzvEI ooxx
= 32
3
23
3)(#### L
zstp
L
zL
fL
zMxuEI o
oyy
cs
w(0) = w(0) = v(0) = v(0) = w(L)= w(L) = v(L) = v(L) = 0
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Integrating Diract Delta and ?ea0iside Function
+
= 32
33)(#### L
zstp
L
zstpf
L
zFzvEI ooxx
+
++++=
+
+
+++=
+ ++=
+
+=
3
2
3
2
3324
33662)(
3
2
3
2
336
3322)(
3
2
3
2
33233)(
3
2
3
2
333)(
44
33
1
2
234
33
22
123
22
12
1
L
zstp
L
z
L
zstp
L
z
f
Lzstp
Lz
FzCzCzCCzvEI
L
zstp
L
z
L
zstp
L
z
f
Lzstp
Lz
FzCzCCzvEI
Lzstp
Lz
Lzstp
Lz
fLzstp
LzFzCCzvEI
Lzstp
Lz
Lzstp
Lzf
LzstpFCzvEI
o
oxx
o
oxx
ooxx
ooxx
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Integrating Diract Delta and ?ea0iside Function
= 32
3
23
3)(#### L
zstpL
zL
fLzMzuEI ooyy
++++=
+++=
++=
+=
3
2
3
2
40
33262)(
3
2
3
2
8332)(
3
2
3
2
23)(
3
2
3
2
2
3
3)(
5
231
22
34
42
123
3
12
2
1
Lzstp
Lz
L
f
Lzstp
Lz
MzCzCzCCzuEI
Lzstp
Lz
L
fLzstp
LzM
zCzCCzuEI
LzstpLzL
fLzstpMzCCzuEI
Lzstp
Lz
L
fLzMCzuEI
o
oyy
ooyy
ooyy
ooyy