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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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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)
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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)
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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:
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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:
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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≈∴∫
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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:
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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
=
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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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−:=
The modulus-weighted shear flow (first moment of area about z-axis) is
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
=
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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⋅:=
The modulus-weighted shear flow (first moment of area about z-axis) is
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:
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
19
SHEAR CENTER:
Calculate the resultant shear forces in each laminate:
VR10
s1
sv1NXY1 sv1( )⌠⌡
d:= VR1 7.165− Nt=
VR20
s2
sv2NXY2 sv2( )⌠⌡
d:= VR2 50− Nt=
VR30
s1
sv3NXY3 sv3( )⌠⌡
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=
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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!
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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)
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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),
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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)
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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)
ME 7502 Lecture 8 Transverse Shear & Torsion in Beams
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)