review 1- bending open and closed sectio

7/25/2019 REVIEW 1- Bending Open and Closed Sectio

1/28

AERSP 301Bending of open and closed section beams

Dr. Jose Palacios


2/28

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


3/28

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 *.


4/28

(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.


5/28

$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!


( )

=

=

=

0

0

l

llz

E

E zzz ==


6/28

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

====


7/28

# the beam is in p&re bending+ aial load res&ltant 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


$hen

$he direct stress becomes!

== AA z dAdA 00

cossin yx +=

( )

cossin yxEE

z +==


8/28


oment Res&ltants!

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


9/28


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


10/28


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+=


11/28

&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


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


12/28

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

%

&


13/28

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


14/28

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)

%

&


15/28

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


16/28

Deflections d&e to bending 'contGd*

&rther+

Beca&se


17/28


#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


18/28


#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


19/28


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


20/28


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)


21/28

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


22/28

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


23/28

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


24/28

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$


25/28

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


26/28

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


27/28

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


28/28

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

review 1- bending open and closed sectio

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