TRANSVERSE SHEAR AND TORSION OF THIN-WALLED …vucoe.drbriansullivan.com/wp-content/uploads/Lecture-8-Part-2... · transverse shear and torsion of thin-walled ... coordinates y,z

ME 7502 Lecture 8 Transverse Shear & Torsion in Beams

1

TRANSVERSE SHEAR AND TORSION OF THIN-WALLED COMPOSITE BEAMS

• REVIEW – TRANSVERSE SHEAR IN SYMMETRIC HOMOGENEOUS BEAMS

• TRANSVERSE SHEAR STRESSES IN THIN COMPOSITE CROSS-SECTIONS

• APPROXIMATE TRANSVERSE SHEAR DEFLECTIONS IN SYMMETRIC CROSS-SECTIONS

• TORSION OF THIN-WALLED OPEN SECTIONS– BEAM RESULTANT LOAD/STRAIN RELATIONS– LAMINATE STRESSES DUE TO APPLIED TORQUE

• TORSION OF THIN-WALLED CLOSED SECTIONS• EXAMPLES• Appendix - ST. VENANT TORSION OF BEAMS, MEMBRANE ANALOGY


2

REVIEW – TRANSVERSE SHEAR STRESS IN SYMMETRIC HOMOGENEOUS BEAMS

TRANSVERSE SHEAR STRESSES:

TRANSVERSE SHEAR FORCE:Mz Mz + dMz

dx

V

V

0)( =+++− dxVdMMM yZZZ

dxdMV Z

y −=∴

σx

σx + dσx

dx

τxyt(y)dxC.A.

cy

ydytyytIdxdMydyt

dxyxd

yt

ydytyxddxyt

c

yz

c

yx

xy

c

y xxy

′′′−=′′′

=

′′′=

∫∫

∫)(

)(/)(),(

)(1

)(),()(

στ

στ

)()(

ytIyVQ

zxy =τ AdyydytyyQ

c

y

c

y′′=′′′= ∫∫ )()(SHEAR FLOW


3

TRANSVERSE SHEAR STRESSES IN THIN-WALLED LAMINATED BEAMS

( ) dxsNdxsNdsdNNdsNF XY

s

s

s

sXYXXXx )()(0 21

2

1

2

1

+−++−==∑ ∫ ∫

dsx

sxNsNsNs

s

XXYXY ∫ ∂

∂=−∴

2

1

),()()( 21

1. COORDINATES y,z ARE PRINCIPAL AXES OF INERTIA (Pyz = -Iyz = 0)2. ARC LENGTH COORDINATE IS s3. EFFECTIVE MODULUS E* AND LAMINATE THICKNESS h ARE FUNCTIONS OF s

ELEMENT FROM BEAM:

x

y

z

Vz

Vy

ss1

s2

Nxy(s2)

Nxy(s1)

Nx(x,s)

Nx(x,s)+dNx(x,s)

h(s1)

h(s2)


4

FROM BENDING EQUATIONS,

( ) ( ) ∫∑∫∑+=−

2

1

2

1

)()()()()()()()( **

**21

s

sx

iziXi

ys

sx

iyiXi

zXYXY dssyshsE

IEV

dsszshsEIE

VsNsN

( ) ( )

−=

∑∑i

ziXi

z

iyiXi

yxX IE

syxMIE

szxMshsEsxN **

* )()()()()()(),(

dxdMV Z

y −= dxdM

VAND yz =

COMBINING,

1. UP TO NOW, EQUATION COMPLETELY GENERAL (IN PRIN MOM OF I AXES)2. IF KNOW NXY = 0 AT ONE POINT, CAN FIND NXY AT ANY OTHER POINT3. EQUATION IS LINEAR IN Vy, Vz , SO CAN CONSIDER EFFECTS SEPARATELY AND

SUPERPOSE.4. PROCEDURE IS VERY COMPLICATED IF CLOSED, NONSYMMETRIC SECTION.

THEREFORE, WILL CONCENTRATE ON OPEN SECTIONS AND ON SYMMETRIC CLOSED SECTIONS. FOR THE LATTER, CAN ALWAYS FIND NXY(s2) = 0


5

OPEN SECTIONS AND CLOSED SYMMETRIC SECTIONS UNDER Vy ONLY (Vz IS IDENTICAL AND CAN SUPERPOSE)

( ) ∫∑=

2

1

)()()()( **1

s

sX

iziXi

yXY dssyshsE

IEV

sN

WHERE s2 IS A LOCATION WHERE NXY = 0

OPEN SECTIONCLOSED SYMMETRIC

SECTION

VyVy

NXY = 0NXY = 0


6

THE TERM IS A YOUNG’S

MODULUS-WEIGHTED SHEAR

FLOW ( ), SO EQUATION CAN BE

WRITTEN

LET hds = dA:

( ) ∫∑=

2

1

)()()( **1

s

sX

iziXi

yXY dAsysE

IEV

sN

∫2

1

)()(*s

sx dAsysE

∑ yAEx*

( )∑∑=

iziXi

XyXY IE

yAEVsN *

*

1)(

ds

s

dA

h

z

y(s)


7

hEEh

o

** = dA

EEdA

o

** = ∫= dAy

EEI

oz

2*

* ∫= ydAEEQ

o

**

*

*

Z

yXY I

QVN =

( )( )

−

−+=

+

−

=

=

∑∑ 1

22*1

1

*

1

* 2 jj

jjjjj

kkkkn

kzk

yXY yy

hsyyEAyE

IE

VN

k

j

FOR PIECEWISE CONSTANT PLATES, IN THE jth PIECE,

THEN

1+jyy

jy

1−jy1y

c

1s

1−js1h

jh

js

NXY = 0 HERE

LET E0 BE A REFERENCE EFFECTIVE COMPOSITE MODULUS (A MODULUS OF ONE OF THE CONSTITUENT LAMINATES), AND


8

FOR OPEN SECTIONS, SHEAR FORCE PER UNIT LENGTH IS ALWAYS ZERO AT ENDS OF SECTION:

- USUALLY EASIEST TO START FROM TOP (SIGN OF EA MOMENT CAN BE NEGATIVE)

OPEN SECTION

Vy

NXY = 0

FOR CLOSED SECTIONS, SHEAR FORCE PER UNIT LENGTH IS ZERO ALONG AXES OF SYMMETRY:

- USUALLY EASIEST TO SPLIT VY OR VzINTO TWO AND TREAT AS OPEN SECTION,

- START FROM “TOP”

Vy

NXY = 0 FOR Vy

NXY = 0 FOR Vy

NXY = 0 FOR Vz

NXY = 0 FOR Vz

Vz

Vz

NXY = 0


9

APPROXIMATE TRANSVERSE SHEAR DEFLECTIONS IN THIN-WALLED LAMINATED BEAMS WITH SYMMETRIC CROSS-SECTION

VERTICAL LAMINATE UNDER TRANSVERSE SHEAR FORCE PER UNIT LENGTH, NXY

∫∫ ≈≈S

XY

XYS

XYV ds

hGsN

Sdss

Sdxdy

0*

0

)(1)(1 γ

LET yV BE VERTICAL DEFLECTION OF PLATE. APPROXIMATESLOPE OF DEFLECTION CURVE BY AVERAGE SHEAR STRAIN IN PLATE:

x

y

z

X

Y

dyV

s

S

dx

NXY

hGsNsor

hGyNy

XY

XYXY

XY

XYXY **

)()()()( == γγ

dxdv

xv

yu

XY =∂∂

+∂∂

=γ

s IS VARIABLE ALONG PLATE MIDDLE SURF.

FROM LAMINATED PLATE THEORY,

FROM STRAIN-DISPL. REL’NS,

∫≈∴S

XYXY

V dssNShGdx

dy

0* )(1∫≈∴

S

XY

XYV dsshsG

sNSdx

dy

0* )()(

)(1VARIABLE PROPERTIES: CONSTANT PROPERTIES:


10

HORIZONTAL LAMINATE:

LAMINATE AT ANGLE θ TO y AXIS, VERTICAL (y) DEFLECTION ONLY:

SINCE NXY IS IN z DIRECTION, THERE IS NO SHEAR DEFLECTION DUE TO NXY

NXY

X

Y

x

y

z

y

z

θ

dyv

dY0

dY0 = dyv.cosθ

S

FROM PREVIOUS PAGE, AVG SHEAR STRAIN IS

∫≈S

XY

XY dsshsG

sNSdx

dY

0*

0

)()()(1

FROM GEOMETRY (SEE FIG.),

θcos10

dxdY

dxdyV =

∫≈∴S

XY

XYV dsshsG

sNSdx

dy

0* )()(

)(cos1

θ ∫≈∴S

XYXY

V dssNhGSdx

dy

0* )(

cos1θ

VARIABLE PROPERTIES: CONSTANT PROPERTIES:


11

NL LAMINATES, EACH WITH CONSTANT PROPERTIES MAKING VARIOUS ANGLES OF θkWITH y-AXIS

ALL LAMINATES HAVE THE SAME DEFLECTION –(COMPATIBILITY)

S1

S2

s1

s2

θ1

θ2

GXY1

GXY2

VY1

VY2

∫≈iS

iiXYiXYiiii

V dssNGhSdx

dy

0* )(

cos1θ

EQUILIBRIUM BETWEEN NXY AND V WITHIN ith LAMINATE:

dxdyGhSdssNV V

XYiiii

S

iiXYiYi

i*

0

cos)( θ=≈ ∫

dxdyGhSdssN V

XYiiii

S

iiXYi

i*

0

cos)( θ≈∴∫


12

VALID FOR CROSS-SECTIONS WHICH ARE SYMMETRIC ABOUT THE y-AXIS. SIMILAR DERIVATION CAN BE MADE FOR DEFLECTION DUE TO Vz

∑=

≈LN

iiXYiii

yV

GhS

Vdxdy

1

2* cos θ

TOTAL SHEAR FORCE IN y DIR’N Vy

=

=

∑

∑

=

=

L

L

N

ii

VXYiii

N

iiYiy

dxdyGhS

VV

1

2*

1

cos

cos

θ

θ

SOLVING FOR dx

dyV


13

EXAMPLE:EXAMPLE

A channel section is made from SP-250-S2 glass/epoxy with [02/45/90/-45]s layup in the flange and a [45/90/-45]s layup in the web as illustrated. The beam is 440 mm long and has a -50 N (downwards) shear force applied in the y-direction at the centroid.

Determine the transverse shear membrane forces per unit length in each section

The centroidal location was found in a previous example, as were the effective constituent elastic constants


14

areas of each laminate:A1 h1 s1⋅:= A1 75mm2=ybar1 56.25 mm⋅:= ybar2 0 mm⋅:= ybar3 56.25− mm⋅:=A2 h2 s2⋅:= A2 135mm2=

momemts of inertia abou t z-axis of individual laminates: A3 A1:= A3 75mm2=

Iz1 A1 ybar12⋅:= Iz1 2.373 105× mm4= Iz3 Iz1:= modulus-weighted moment of inertia of section:

Iz2112

h2⋅ s23⋅:= Iz2 1.424 105× mm4= EIz E1 Iz1⋅ E2 Iz2⋅+ E1 Iz3⋅+:=

EIz 1.442 1010× MPa mm4⋅=Iz3 A3 ybar32⋅:= Iz3 2.373 105× mm4=

SOLUTION:

Let top flange be laminate no. 1, web be laminate no. 2, and bottom flange be laminate no. 3.

a. Use the following form of the shear force per unit length equation:

MPanewton

mm2:= Nt newton:=Unit definitions:

Young's modulus thickness section lengthLaminate properties:

laminates 1 and 3: E1 26.12 103⋅ MPa⋅:= h1 2.00 mm⋅:= s1 37.5 mm⋅:=Shear force: Vy 50− Nt⋅:=

laminate 2: E2 14.20 103⋅ MPa⋅:= h2 1.20 mm⋅:= s2 112.5 mm⋅:=

y- distances from section centroid to individual laminate centroids:


15

In Laminate 1: let s1v be a variable that runs from 0 to s1 along the middle surface of laminate 1. Let ybarv1 be the cordinate from the centroidal z axis to the centroid of the portion of area contained in s1v .

ybarv1 56.25 mm⋅:=

The modulus-weighted shear flow (first moment of area about z-axis) is

EAybarv1 sv1( ) E1 sv1⋅ h1⋅ ybarv1⋅:=

N.xy in laminate 1 isNXY1 sv1( ) Vy

EAybarv1 sv1( )EIz

⋅:=

Plot along top of beam: sv1 0.mm .1 mm⋅, s1..:= Sv1 sv1( ) s1 sv1−:=

0 20

0.3

0.2

0.1

0

NXY1 sv1( )Nt

mm

Sv1 sv1( )mm

NXY1 0 mm⋅( ) 0Ntmm

=

NXY1 s1( ) 0.382−Ntmm

=


16

In Laminate 2: s2v is variable from 0 to s2 along the middle surface of laminate 2. y barv2 is coordinate from the centroidal z axis to the centroid of the portion of area contained in s2v .

ybarv2 sv2( ) s2

2

sv2

2−:=


EAybarv1 E1 s1⋅ h1⋅ ybarv1⋅:=

EAybarv2 sv2( ) E2 sv2⋅ h2⋅ ybarv2 sv2( )⋅:=

N.xy in laminate 2 is NXY2 sv2( ) VyEAybarv1 EAybarv2 sv2( )+

EIz⋅:=

Plot along web of beam: sv2 0.mm .1 mm⋅, s2..:= y2 sv2( ) s2

2sv2−:=

0.6 0.4 0.2 050

0

50

y2 sv2( )mm

NXY2 sv2( )Nt

mm

NXY2 0 mm⋅( ) 0.382−Ntmm

=

NXY2s2

2

0.476−Ntmm

=

NXY2 s2( ) 0.382−Ntmm

=


17

NXY3 s3( ) 0Ntmm

=

NOTE:1. NXY = 0 WHERE IT SHOULD BE2. NXY IS CONTINUOUS AT LAMINATE JOINTS

NXY3 0 mm⋅( ) 0.382−Ntmm

=

0 200.4

0.2

0

NXY3 sv3( )Nt

mm

sv3mm

sv3 0.mm .1 mm⋅, s3..:=Plot along bottom of beam:

NXY3 sv3( ) VyEAybarv1 EAybarv2+ EAybarv3 sv3( )+

EIz⋅:=N.xy in laminate 3 is

EAybarv2 E2 s2⋅ h2⋅ 0⋅ mm⋅:=

EAybarv3 sv3( ) E3 sv3⋅ h3⋅ ybarv3⋅:=EAybarv1 E1 s1⋅ h1⋅ ybarv1⋅:=


ybarv3s2

2−:=E3 E1:=s3 s1:=

h3 h1:=

s3v is variable from 0 to s3 along the middle surface of laminate 3. y barv3 is coordinate from the centroidal z axis to the centroid of the portion of area contained in s3v .

In Laminate 3:


18


19

SHEAR CENTER:

Calculate the resultant shear forces in each laminate:

VR10

s1

sv1NXY1 sv1( )⌠⌡

d:= VR1 7.165− Nt=

VR20

s2


d:= VR2 50− Nt=

VR30

s1


d:= VR3 7.165− Nt=

Torque is not zero!! Tx VR1 s2⋅ VR2 12.59⋅ mm⋅+( )−:= Tx 1.436 103× Nt mm⋅=

To counterbalance the couple set up by transverse shear, torsional stresses occur in beam:

In order to prevent torsional stresses and rotation of cross-section, mustapply V at an offset location called the SHEAR CENTER, d:

dTx

Vy:= d 28.711mm=


20

TORSION OF THIN-WALLED OPEN SECTIONS

SINGLE PLATE, M/P SYMMETRIC LAMINATE – KNOW THAT Tx = -sMXY

0=== XYYX NNN φκ ′−=XY 0== YX MM

′−

=

φκκ

Y

X

ttt

t

t

XY DDDDDDDDD

M 666261

262212

161211

00

( )( )

( ) ( )φκ

φκκ

φκκ

′−=−

=′−+

=′−+

ttY

tYX

tYX

DDDDDDDDDDDDDDD

2611161222112

12

26221211

16121112

0

0

{M} = [D]{κ} IS 3 EQNS WITH 3 UNKNOWNS: κx, κy, κxy

FROM 1st 2 EQNS,

SUBTRACT:

ALSO KNOW,

ANGLE CHANNEL J-STIFFENER I-BEAM T-BEAM

Txx

y

X

Y

s

-φ’LL

WHERE φ’ IS BEAM ANGLE OF TWIST PER UNIT LENGTH


21

SIMILARLY, φκ ′−

−=

22112

12

16222612

DDDDDDD tt

X

φκκ ′−+= ty

tx

tXY DDDM 666261

( ) ( ) φ′

+

−−+−

−=∴ ttttttt

XY DDDD

DDDDDDDDDDM 662122211

26111612621622261261

sD

. NOW SINCE , EQNS FOR STRESS AND DEFORMATION

OF PLATE-BEAM BECOMEφ′−=∴ sXY DM sMT XYx −=

s

xxXY Ds

Ts

TM =′−=∴ φ,

φκ ′−

−=∴

22112

12

26111612

DDDDDDD tt

YGET:

3rd EQUN IS: SUBBING FOR κx AND κy,

CALL THIS


22

SECTION COMPOSED OF JOINED MULTIPLE PLATES RESTRICTING X-PLATE CURVATURE

0,0 =≠ XXM κ

IN EACH SECTION,

COMPATIBILITY:

0,0 ==iXYiM κ

φκ ′−=iXY

{M} = [D]{κ} BECOMES:

′−

=

φκ

i

iii

iii

iii

i

I

Yttt

t

t

XY

X

DDDDDDDDD

M

M 00

666266

262112

161211

φ′

−= t

t

X i

i

i

iiD

DD

DM 1622

2612

φ′

−= t

tt

XY i

i

i

iiD

DD

DM 6622

2662φκ ′=

i

i

i DDt

Y22

26

THIS IS 3 EQNS IN 3 UNKNOWNS FOR A GIVEN φ’ – MX, MXY, κY

SOLVING, GET


23

−=

i

ii

ii DDD

DDtt

tm

22

266266 φ′−=

ii mXY DM

APPLIED TORQUE MUST BE IN EQUILIBRIUM WITH SUM OF COUPLES ON EACH CONSTITUENT PLATE. IF WE LET

φ′+=−= ∑∑ kmkXYx sDsMTkk

∑+=′∴

kkm

x

sDT

k

φ

∑−=∴

kkm

xmXY sD

TDM

k

i

i ∑

−=

kkm

xtt

X sDTD

DDD

Mk

i

i

ii

i 1622

2612

STILL NEED TO FIND RELATION BETWEEN φ’ AND TORQUE Tx:

, THEN

AND TOTALTORQUE Tx BECOMES:

FROM WHICH:

AND

NOTE THAT MX IS NOT ZERO!


24

TORSION OF THIN-WALLED CLOSED SECTIONS (TUBES)

φ′ = ANGLE OF TWIST/UNIT LENGTH OF BAR = CONST.

r, θ ARE POLAR COORDINATESTx

Txφ’L

LWHEN TORQUE IS APPLIED, CROSS-SECTION:

- ROTATES DUE TO SHEAR STRAIN γXY = 2εXY ONLY

- WARPS (PLANE SECTIONS DO NOT REMAIN PLANE, BUT MAKES VARIABLE ANGLE α WITH ORIGINAL SECTION)

,0==== YXzy NNMMSINCE 0==oo YX εε

THEREFORE, ONLY 0XYXY ANDN ε WILL BE NON-ZEROX

iXYτ

IN A LAYER: 0=nXτ

0≈nXτ (THIN WALL)

n

x

n

x

IN LAMINATE:

WHY IS THERE NO MX HERE?

0≈== XYYX κκκ


25

INTERNAL EQUILIBRIUM:

00

=∴

−−+==∑XY

XYXYXYX

dNdNNNF

dsXYN

dSdS

dNN xy

XY +

X

…OR OF A FINITE SECTION:

2XYN

1XYN

OF AN ELEMENT…

21 XYXY NN =

constant=XYNEITHER WAY, (T.1)


26

FROM FIGURES,

θβ rdds =cos (T.3)

θ

α

−γXY

rφ’dx

α

−γXY

x,X

Y

rφ’dxcosβ

α+(−γXY)

r

rφ’dx

β

φ’dx

LOOKING DOWN FROM TOP:

GEOMETRY OF DEFORMATION:

βφβφγα coscos ′=′

=− rdx

dxroXY

(T.2)

COMPATIBILITY:

αds IS AMOUNT THAT CROSS SECTION WARPS, i.e., POINTS ALONG CROSS SECTION MOVE PARALLEL TO X-AXIS. SINCE TOTAL MOTION MUST BE ZERO IF ONE GOES COMPLETELY AROUND SECTION (NO “SPLIT” ALLOWED ALONG x-AXIS), MUST HAVE

0=∫sdsα (T.4)∫∫ ′−=

ss XY dsrds βφγ cos

oXYr γβφα +′=∴ cos

USING (T.1),


27

( ) ,cos rdSNdT XYx ⋅−= β

CONSTANT FROM (T.1)

rdθ FROM GEOMETRY

EQUN (T.3)

mXYmXYx ANdANT 22 −=−=∴ ∫

))((21 rrddAm θ=

m

XXY A

TN2

−=∴

y

zr

ds

β

EXTERNAL EQUILIBRIUM:

NXY AROUND CIRCUMFERENCE MUST EQUILIBRATE APPLIED TORQUE

XYN∫−=S

XYx dSrNT βcos

∫−=∴ θdrNT XYx2

TO CALCULATE INTEGRAL AND FIND T, LOOK AT AREA Am ENCLOSED BY MIDDLE SURFACE OR MEDIAN THICKNESS LINE:

rdθ

r

dAm

MEDIAN LINE

Am

(T.5)


28

( ) ( ) ( )

( ) ( ) ( )shG

sGsh

shdS

GAT

shsGdS

AT

o

XY

Som

x

S XYm

x

*

**

**2*2 44

=

==′ ∫∫φ

)()()( * shsG

NsXY

XYXY =∴γ

NOW TO GET IN TERMS OF T:φ′

FORCE/STRAIN EQN (T.6) INTO COMPATIBILITY EQN (T.3) GIVES

( ) ( ) ( ) ( )∫∫∫∫ −=′∴′=′=′=−S XYm

XYmm

XY

XY

shsGdS

ANAdAdSr

shsGdsN

** 2,22cos φφφβφ

USING (T.5),

WHERE

OR, SINCE ,2 66* tXY AhG =∗

( )∫=′S

tm

x

sAdS

AT

6622

φ

MIDDLE SURFACE SHEAR FORCE/SHEAR STRAIN EQUN:

,)()()(* sshsGN XYXYXY γ= (T.6)

(T.7)


29

CASE OF TUBE MADE OF ONLY ONE LAMINATE:

THICKNESS h AND EFFECTIVE SHEAR MODULUS G*XY ARE CONSTANT:

m

xXY A

TN2

−= (NXY UNAFFECTED BY ANYTHING BUT Am)

tm

x

XYm

x

As

AT

hGs

AT

662*2 24

==′φ

CASE OF ONE TUBE WITH MULTIPLE SECTIONS EACH WITH CONSTANT h & A66 (OR G*XY ):

∑∑ ==′i i

ti

m

x

i iXYi

i

m

x

As

AT

Ghs

AT

662*2 24

φ

m

XXY A

TN2

−= (NXY UNAFFECTED BY ANYTHING BUT Am . NOTE, HOWEVER, STRESSES IN SECTIONS WILL NOT BE SAME!)

(Am IS AREA CONTAINED INSIDE MEDIAN LINE OR MIDDLE SURFACE)

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TRANSVERSE SHEAR AND TORSION OF THIN-WALLED …vucoe.drbriansullivan.com/wp-content/uploads/Lecture-8-Part-2... · transverse shear and torsion of thin-walled ... coordinates y,z