bending and transverse tension

8/10/2019 Bending and Transverse Tension

1/23

BENDING AND

TRANSVERSE SHEAR


2/23

TRANSVERSE SHEAR OF BEAMS

definition

side panels of a beam element are subjected to shearing forceand a perpendicular vector component of bending moment

at the same instance.

y

xz

MxVy

2/23

y

xz

dz lC

C

Vy

MxMy

Assumptions:

straight beam axis

???

homogeneous material, oo!e"s la# applies

$ constant cross section:A%z& 'A, Sx%z& ' Sx,Ix%z& 'Ixetc.


3/23

3/23

Reminder

stress formula for simple uniaxial bending

simple shear

complementarit( of shear stresses

z

dz

yzzy =

Mx

I xy

Mx

Vy

y

xzVy

z zy=Vy

A

uniform stress distribution

%rough approximation onl(&:

yzzy

xzy

zy' yz



4/23

)/23

*ac!groundK

yz

zy

z

yz: +eroat %unloaded& external surfaces

complementarit(: zy, corneris also

!

cross sections do not remain plane %???...&

zy =Mx

I xy holds just approximatel(

yzzy-

yz: longitudinal shear

zy 'V

y

A, but zymax is even larger



5/23

/23

*ac!ground

hear stresses #ithin the plane of the cross sectionat the boundar( are tangential

K

C

dAz

z'

t

tz-

zt-

0 distribution of both directions and magnitudes ofstresses calculated from simple shear are contradictor(

zy =Mx

I xy : can be !ept altogether,

zy%y& % 0zt& ???

so"id sections thin#$a""ed sections



6/23

1/23

45:

hear stresses parallel to the shearing force in solid sections

hear stresses perpendicular to the shearing force in solid sections

6ongitudinal shear of a finite segment of beams #ith solid sections

hear stresses in thin7#alled cross sections loaded in their s(mmetr( axis

hear stresses in thin7#alled cross sections loaded orthogonall( to their s(mmetr(

axis, the concept of shear centre



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8/23

9urther assumptions:

s(mmetric cross section, V' Vyis directed along the s(mmetr( axis,

M'Mx%bending moment vector is perpendicular to the shearing bending is uniaxial&,

normal stresses arise onl( from bending ,

shear stresses arise onl( from shearing.

zy =MxI x y

ANA%&SIS OF THE BEAM E%EMENT

GEOMETRI' e;uations

C

dx

$ rig. c. s.: x= y= xy=,

z%y,z& = y = x%z&ydx%z&

dz

zx%x,y,z& =

du%x,y,z&

dz

zy%y,z& =dv%y,z&

dz

ne more assumption:

vertical displacement v%arising

strictl( from shear& depends onl( ony

but not onx%that is, zyis constant

#ithin a given hori+ontal section&

%;uasi& plane cross sections %for bending&:

TRANSVERSE SHEAR OF SO%ID BEAMS


8/23


9/23

B/23

STATI'A% e;uations for a section of the beam e"ement

C

z

y

xy

z

Mx$ dMx

dz

%%y&

A'

Vy$ dVy

MxVy

yy

&'$d&'&'$d&' &'

d

dz

d

d z%y& dA' yz%y&%%y& dzA'

"#z

: %&'$d&'&d&'' d&''d

the order of differentiationand integration is reversible:

dA' yz%y&%%y&A'dz%y&

dz

dA' yz%y&%%y&A'

dMxdz

y

Ix

Vy%z& 'dM

x%z&

dz

z%y& ' yMx%z&

Ix

ydA' yz%y&%%y&A'

Vy

Ix

Sx' : first moment of the sectionA'about the central axisx

the shear formula:

yzy =zyy=$y Sx

' y

Ix %y

Vy

zy * !



10/23

A'

C/23

zy * !

CMx

Vy

zy=$y Sx

'

Ix %

Vy

A'

zy

C

MxVy

zy y =$y Sx

' y

Ix %

Vy

zy

for sections #ith side#alls parallel to Vy%%is constant&: zy,max: at the maximum of S'x,

i. e., at the height of the centro#(

zy y =$y Sx

' y

Ix % y

Vy

for a generic cross sectionzy,max: at the maximum of S'x/%



11/23

CC/23

zy + zx * !

SMx

Vy

zy=$y Sx

'

Ix %

Vy

A'

zy

zx$

t

)

**

t

zx,max= zytan

zy

max= zt

max=zt=zx , max

2zy2

max=

zy

cos

maximum shear stress

at the t#o boundaries:

t

zy

*zx*

ze*

ze*=zx

*2zy* 2

resultant of theshear stress at a point*:

*



12/23

C2/23

zy + H* ! ,ca"c-"ation of the res-"tant "ongit-dina" shear.

""



13/23

%y&

C3/23

zy + H* !

yz=zy=$y Sx

'

I x %

Vy

C

z

y

x

lA'

dz

d

Vyz

%%y&

l%y& ' d%y,z& ' yz%y,z&%%y& dzzC

z2

d%y,z& 'yz%y,z&%%y& dz

it is alread( !no#n:

KC

K2

moreover,yis fixed:

zC

z2

l' %dzzC

z2Vy%z& Sx'

Ix%

l' Vy%z& dzzC

z2Sx'

Ix AV: area of the shearforce diagram atlength l'z2DzCl' AV

Sx'

Ix


,ca"c-"ation of the res-"tant "ongit-dina" shear.


14/23

C)/23

$ith shearing force /ara""e" to $ith shearing force /er/endic-"ar tothe s0mmetr0 a1is the s0mmetr0 a1is

Vy Vy

MxMx

TRANSVERSE SHEAR OF THIN#2A%%ED BEAMS

9urther assumptions:

M'Mx%bending moment vector is perpendicular to the shearing bending is uniaxial&,

normal stresses arise onl( from bending ,

shear stresses arise onl( from shearing.

shear stresses are parallel to the #all of the section,

shear stresses in a section perpendicular to the #all are constant.

zy =

Mx

I x y


15/23

C/23

z%y,z& '=z%y,z&

zt%t,z& ' !=zt%t,z&

EF

>tzEF >zt z

('

EF

tzEF zt @z

) '

$ rig. c. s.: x=y=xy=,


dz

zn%n,t,z& ' ' den%n,t,z&

dz

zt%t,z& = det%y,z&dz

"#x: zx%x,y,z& dA'Vx%z& '

"#y: zy%y,z& dA' Vy%z&

M#z: %zx%x,y,z&y$

zy%y,z&x& dA' $%z& '

A%z&

A%z&

A%z&

M#x: z%x,y,z&ydA'Mx%z&A%z&

$ith shearing force /ara""e" to the s0mmetr0 a1is

tn


GEOMETRI' e;uationsSTATI'A% e;uations

MATERIA% e;uations

automaticall(satisfieddue to

s(mmetr(

zyis constant in a

hori+ontal plane


16/23

C1/23

Vy

Mx

+

+dz

x+Dx

v

v

v

A'

z

z

xz

z$ dz

x

y

"#z: dz%y& dA' xz%x,y& vdzA'

dA' xz%x,y& vA'

dz%y&

dz

dA' xz%x,y& vA'dM

xdz

y

Ix

ydA' xz%x,y& vA'

Vy

Ix

e;uilibrium of #idth +Dx:

Sx' : as before

approximation: the centroid of area of#idth +Dxis at a height of /2 %vGG &

xzx=zx x=$y Sx

' x

Ix v

Vy

zx/2x=

$y +x v/2

Ix v =

$y +x

2Ix

Vy Vy




17/23

C8/23

zx/2x=

$y +x v/2

Ix v =

$y +x

2Ix

Vy Vy

Vy

flanges: hori+ontal shear stresses onl($

$

zxmax=

$y +

2

Ix

Vy

zx

max

zxmax

Mxzt $

zymax

#eb: vertical shear stresses onl(

zymax=$y Sx

' max

Ix vVy

%the shear formula&zy y =$y Sx

' y

Ix % y

Vy

maximum stressat the centroid:

stress at the top orbottom of the #eb:

zy-

$y 2+v /2

Ix v =

$y +

Ix

$y

zx

max

zy.

$y 2+v /2

Ix v =

$y +

Ix

Vy Vy shear f"o$




18/23

CtzEF >zt z

('

EF

tzEF zt @z

) '

x=

y=

xy=,


dz

zn%n,t,z& ' ' den%n,t,z&

dz

zt%t,z& = det%y,z&dz

"#x: zx%x,y,z& dA'Vx%z& '

"#y: zy%y,z& dA' Vy%z&

M#z: %zx%x,y,z&y$

zy%y,z&x& dA' $%z& ' !

A%z&

A%z&

A%z&

M#x: z%x,y,z&ydA'Mx%z&A%z&

tn


GEOMETRI' e;uationsSTATI'A% e;uations

MATERIA% e;uations

$ rig. c. s.:

satisfied due tos(mmetr(

zyis constant in a

hori+ontal plane

$ith shearing force /er/endic-"ar to the s0mmetr0 a1is


19/23

CB/23

Vy

Mx

z

x

ydz

x+Dx

z

xzz$ dz

xzx=zx x=$y Sx

' x

Ix v

Vy

zx/2x=

$y +x v/2

Ix v =

$y +x

2Ix

Vy Vy

%calculation of stresses: as in sectionss(mmetric abouty&

H#isting D #h(?

Vy




20/23

/=$y +

2

2v

)Ix$'

Vy

2/23

Vy

$zxmax

Mxzt

$

zymax

zxmax

zxmax=$

y

+

2Ix

Vy

zy.

$y +v /2

Ix v =

$y +

2Ix

Vy Vy

/

/

0' Vy

/=zx

max+v

2=

$y +2

v

)Ix/

Vy

C

0' Vy

C,*.

"' Vy

C,*. CS

e

Vy

e=+

2

2v

)Ix



H#isting D #h(?


21/23

zy.

$y +v /2

Ix v =

$y +

2Ix

Vy Vy

2C/23

CS: the shear centre,vertical force"passing through

this point causes no t.#%t#n-

Vy

$zxmax

Mxzt

$

zymax

zxmax

zxmax=$

y

+

2Ix

Vy

"

e

CS



Io t#isting J

/=$y +

2

2v

)Ix$'

Vy

/

/

0' Vy

/=zx

max+v

2=

$y +2

v

)Ix/

Vy

C

0' Vy

C,*.

"' Vy

C,*. CS

e e=+

2

2v

)Ix


22/23

C

22/23

So far333

$

zt

$

,*.C

,*.! "CCS

e xS "' Vy

$'" %e$xS&

Mx 4 Vyand Ttogether$ere considered3

To ha5eMx and Vyon"04 s-/erim/ose 6Ton it ,7.

C

1$,*.

C

"' Vy

C

,*.Vy

Mx



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2C/2C

'riteria

2

bending and transverse tension

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