Post on 22-May-2022
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Lecture Note 3Lecture Note 3
Stress Functions
First Semester, Academic Year 2012,Department of Mechanical Engineering
Chulalongkorn University
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Objectives #1
Describe the equation of equilibrium using 2D/3D stress elements
Set well-posed elasticity problems (e.g. for finite element analyses)S l i l l ti bl b th i d i Solve simple elastic problems by the inverse and semi-inverse method
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Concepts #1
Equation of equilibrium is the governing equation. Well-posed problems can be formulated by substituting Well posed problems can be formulated by substituting
appropriate constitutive relationships, loads, initial conditions and boundary conditions into the governing equationsWith i t diti f tit ti l ti th With appropriate conditions from constitutive relations, the stress distribution can be found by fitting the boundary conditions into the appropriate stress functions
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Equilibrium ElementEquilibrium X, Y, Z are body
forces per unit lvolume
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5
Equilibrium Complementary Principal of ShearEquilibrium
0zM
x x
( )( ) ( )( )
2 2
( )( ) ( )( ) 0
xyxy xy
yx
x xy z x y zx
y yx z y x z
( )( ) ( )( ) 02 2
02 2 2 2 2 2
yx yx
xy yxxy xy yx yx
x z y x zy
V V V V V Vx y
2 2 2 2 2 2
2 2 0
xy xy yx yx
xy yxxy yx
x y
x yx y
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As 0 and 0, , etcxy yx
x yx y .
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Equilibrium 3D EquationsEquilibrium X, Y, Z are body
forces per unit l
0 ( )( ) ( )x
x x xF x y z y zx
volume
( )( ) ( )yxyx yxy x z x z
y
( )( ) ( ) ( ) 0zxzx zxz x y x y X x y z
z
0xyx xz Xx y z
0
0
yx y yz
zyzx z
Yx y z
Z
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0yzx z Zx y z
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Equilibrium 2D Plane StressEquilibrium 2D Reduction
z 0yz zx
z yz zx
xzxyx
x y
0X
zy
yzyx y
x y
0Y
z
zx
zy
x
z
y
0Z
z
0xyx Xx y
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0yx y Y
x y
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Equilibrium 2D Plane StrainEquilibrium 2D Reduction
z const, 0yz zx
z , yz zx
xzxyx
x y
0X
zy
yzyx y
x y
0Y
z
zx
zy
x
0z Zy z
0xyx X
x y
0yx y Yx y
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0z Z
z
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Equilibrium Components on an Inclined PlaneEquilibrium
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Equilibrium Boundary ConditionsEquilibrium 3D Scale Up
0xF
Unit thickness
1 02
x
x yxX s y y X x y
As 0,
x yx
xdy dxXds ds X l
x yx
ds ds
X l m
x yx zx
xy y zy
X l m n
Y l m n
xy yY l m xz yz zZ l m n
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coscos
x
y
lm
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Equilibrium Strain Compatibility #1Equilibrium
2 2 2xy
xyv u v ux y x y x y x x y y
2 2 2
2 2xy
yz
x y x y x y x x y y
w v v uy z x y y xx y
zx
y z x y y xx yu wz x
2 2 2 2 2
2 22
xy y x
yz y
x y x y
2 2
2 2 2
yz yz
zx x z
y z y z
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2 2z x z x
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Equilibrium Strain Compatibility #2Equilibrium
2 2 2 2
( ) ( )xy zx v u u wz x x y z x x y x y z x
2 2 2
2( ) ( )xy zx
z x x y z x x y x y z xu v w u
x z y z x y z y x y zx
y y y y
2
2 ( )yz xyx zx
2
2 ( )
2 ( )
yz xyx zx
y yz xyzx
y z x x y z
2
2 ( )
2 ( )yz xyz zx
z x y x y z
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( )
x y z x y z
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Equilibrium Strain Compatibility #3Equilibrium
2 2 2
2 2xy y x
x y x y
2 2
2 22
2 2yz yz
x y x y
y z y z
2 2 2
2 2zx x z
y z y z
z x z x
2
2
2 ( )yz xyx zx
y z x x y z
2
2
2 ( )y yz xyzx
z x x x y z
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2 ( )yz xyz zx
x y x x y z
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Equilibrium 2D Strain CompatibilityEquilibrium 2D Reduction
z z0 or const, 0yz zx yz zx
2 2 2
2 2
2
xy y x
x y x y
2
2 x
y z
(yz
x
zx
x
xy
y z)
2yz
2
2z
y z y
2
2y
z2
2
2 y
z x
(yz
x
zx
x
xy
y z)
2zx
2
2x
z x z
2
2z
x
2
2 z
x y
(yz
x
zx
x
xy
y z)
2 2 2xy y
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2 2xy y x
x y x y
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2D Elastic Problems 3D Set Up #1 Problem
Governing equations
0xyx xz Xx y z
0yx y yz Yx y z
0zyzx z Z
x y z
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2D Elastic Problems 3D Set Up #2 Problem
Constitutive equations
1
xuxv
1 ( )
1 ( )
x x y zE
y
z
yw
( )
1 ( )
y y z x
z z x y
E
E
z
xy
zu vy x
y
xyxy
E
G
yz
yv wz y
yzyz G
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zxw ux z
zxzx G
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2D Elastic Problems 3D Set Up #3 Problem
2 2 2
2 2xy y x
Compatibility condition
2 2
2 22
2 2yz yz
x y x y
y z y z
2 2 2
2 2zx x z
y z y z
z x z x
2
2
2 ( )yz xyx zx
y z x x y z
2
2
2 ( )y yz xyzx
z x x x y z
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2 ( )yz xyz zx
x y x x y z
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2D Elastic Problems Plane Stress Set Up #1 Problem
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Governing equations
Constitutive equations
1 ( )
1 ( )
x x yE 0xyx X
x y
( )
1 2(1 )
y y x
xy xy xy
E
G E
0yx y Yx y
xy xy xyG E
2 2 X
2
2
2 2
0xy xy xyx x x XX Xx y y x x y xx
Y
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20yx y xy y xy y YY Yx y x y x y yy
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2D Elastic Problems Plane Stress Set Up #2Problem
Compatibility condition2 2
2 2xy x X
2 2 2
2 2
2 2 2
xy y x
x y x y
2
2 2
2xy y
x y xx
Y
2 2 2
2 2
2 2 22 2
2(1 ) ( ) ( )
2
xyy x x yx y x y
2x y yy
2 2 2 2
22
2(1 )
(1 )( )
xy y yx x
yx
x y x x y y
X Y
2 2
2 22 2
(1 )( )yx
y yx x
x yx y
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2 2 2 2x x y y
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2D Elastic Problems Plane Stress Set Up #3Problem
2 2 22 2 2
2 2 2 2 2 2(1 )( )y y yx x xX Yx yx y x x y y
2 2 2 2 2 2
2 22 2
2 2 2 2 (1 )( )y yx x
x yx y x x y y
X Yx yx y x y
2 22 2
2 2 2 2y yx x
x yx y x y
x x y y
2 22 2
2 2 2(1 )( )y yx x
y y
X Yx yx y x y 2
2 22 2
2 2 2 2
2 22 2
(1 )( ) y yx xX Yx y x y x y
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2 22 2
2 2 2 2(1 )( ) y yx xX Yx y x y x y
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2D Elastic Problems Plane Stress Set Up #4Problem
Compatibility condition
2 22 2
2 2 2 2(1 )( ) y yx xX Yx y x y x y
x y x y x y
2 2
2 2( )( ) (1 )( )x yX Y
2 2( )( ) ( )( )x y x yx y
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2D Elastic Problems Plane Stress Set Up #5Problem
Neglect the body force
2 2
2 2( )( ) 0x yx y
0xyx
x y
x y 2 2
( )( ) 0
xy x
y x
2 2
2 2 2 2
( )( ) 0
( )( ) 0
x yx y
0yx y
x y
2 2 2 2
4 4 4 4
2 2 4 4 2 2
( )( ) 0
0
x y y x
Assume a function
xy y
x y
2 2 4 4 2 2
4 4 4
4 2 2 42 0
x y x y y x
x x y y
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2 2 2
2 2, , x y xy x yy x
x x y y
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Stress Function Airy Stress Function Airy
Neglect the body force2 2
( )( ) 0
To solve the equations use stress functions that satisfy
2 2( )( ) 0x yx y
To solve the equations, use stress functions that satisfy relationships
2 2 2
Substitute into the compatibility condition
2 2, , x y xy x yy x
Substitute into the compatibility condition
4 4 4
4 2 2 42 0x x y y
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x x y y
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Inverse MethodPlate
Procedure Select the stress function Select the stress function Check compatibility Fit in the boundary condition
AdvantageSimplified assumptions Simplified assumptions
Disadvantage Find problems that fit solution
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Example Timoshenko 1 #1 Plate
For an elastic plate subjected to static loads shown, determine the stress function. The A, B and C are constantsconstants.
Domain &
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Boundary conditions
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Example Timoshenko 1 #2 Plate
Describe stress distributions within the platewithin the plate
2 2
4 4 4 4 4 4
4 2 2 4 4 2 2 40, 0, 0 2 0
Ax Bxy Cy
x x y y x x y y
4 2 2 4 4 2 2 4
2 2 2
2 22 , 2 , x y xy
x x y y x x y y
C A Bx yy x x yy x
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Example Timoshenko 1 #3 Plate
2 2, 2 , 2 , x y xyAx Bxy Cy C A B
Let 1, 2, 3
6 MPa, 2 MPa, 2 MPa x y xy
A B C
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Example Timoshenko 1 #4 Plate
6 MPa, 2 MPa, 2 MPa x y xy
Plane stress, 0 MPa
6 83 MP
z
1
2
6.83 MPa 1.17 MPa 0 MPa
3
1 3
0 MPa
6.83 MPaT
1 3
2 2 21 2 2 1( ) ( )
2
T
v
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6.32 MPav
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Example Timoshenko 2 #1 Plate
For an elastic plate subjected to static loads shown, determine the stress function. The D is a constant.
Boundary conditionsFor left & right edges: 0Dy
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For left & right edges: , 0
For top & bottom edges: 0, 0x xy
y xy
Dy
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Example Timoshenko 2 #2 Plate
Describe stress distributions
3 2 2 3
6 2 2 6A B C Dx x y xy y
4 4 4
4 2 2 4
4 4 4
0, 0, 0x x y y
4 4 4
4 2 2 42 0x x y y Satisfy the condition if
F l ft & i ht d
2
2x Cx Dyy
For left & right edges
x C
For top & bottom edgesx Dy Dy
2
2
2
y Ax Byx
For top & bottom edges
y A x B 0
For all edges
y
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2
xy Bx Cyx y
For all edges
xy B x C 0y
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Example Timoshenko 2 #3 Plate
0 0Dy
, 0, 0Let 1 MPa/m, 2 m, 1 m
1 , 0, 0
x y xy
x y xy
Dy
D a byx y xyy
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Example Timoshenko 3 #1 Plate
For an elastic plate subjected to static loads shown, determine the stress function. The B is a constant.
State the domain & boundary conditionsconditions
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Example Timoshenko 3 #2 Plate
4 3 2 2 3 4
12 6 2 6 12A B C D Ex x y x y xy y
4 4 4
4 2 2 42 , 2 , 2A C Ex x y y
4 4 4
4 2 2 42 0
2 4 2 0x x y yA C E 2
2 2
2 4 2 0Thus, (2 )A C E
E C A
2 22
2 2(2 )
x Cx Dxy EyyCx Dxy C A y
22 2
2
2
y Ax Bxy Cyx
B D
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22 22
2 2xyB Dx Cxy y
x y
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Example Timoshenko 3 #3 Plate
Describe stress distributions
2 2(2 )x Cx Dxy C A y
22 2
2
2
y Ax Bxy Cyx
B D
22 22
2 2xyB Dx Cxy y
x y
Satisfy the condition if
x C 2x D (2xy C A )y
y A 2x Bxy C 2
2 2
y
B x C Dxy 2y
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22xy x C xy
2y
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Example Timoshenko 3 #4 Plate
2
Let 0, 2 MPa0 MPa 2 MPa MPaA C D B
xy x 20 MPa, 2 MPa, MPax y xyxy x
2 MPay xy 2 MPaxy x
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y
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Example Timoshenko 3 #5 Plate
2 2 2 2
Let 1 MPa, 2 MPa, 3 MPa, 4 MPa3 4 7 MPa 2 3 MPaA B C Dx xy y x xy y
2 2
3 4 7 MPa, 2 3 MPa,
6 2 MPax y
xy
x xy y x xy y
x xy y
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x y xy
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Semi-Inverse MethodBeam
Procedure Select the stress function by assuming reasonable Select the stress function by assuming reasonable
assumptions Check compatibility Fit in the boundary condition
Advantage Advantage More versatile stress functions
Disadvantage Inaccuracies in regions which assume the St. Venant’s
principle
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Example Beam 1#1 Beam
For an elastic plate subjected to static loads shown, determine the stress distribution and displacements. Assume unit thicknessAssume unit thickness.
State the domain & boundary
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boundary conditions
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Example Beam 1 #2Beam
3BAxy xy
4 4 4
4 2 2 4
6
0, 0, 0
y y
x x y y
4 4 4
4 2 2 42 0
x x y y
x x y y
2 2 22
2 2, 0, 2x y xyBBxy A y
x yy x
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Example Beam 1 #3Beam
2 2 22
2 2, 0, 2x y xyBBxy A y
x yy x
2
Top and bottom edges BC: 0 at 2xy
x yy xhy
2 2
2
02 4 8B h BhA A
Bh B
2, 0, 8 2
Right edge BC: at 0
x y xy
xy
Bh BBxy y
P dA x
2 3/ 2 2
/ 2( )
8 2 12
y
h
h
Bh B BwhP y wdy BI /B P I
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2 2 2 22 2
12 3, 0, ( 4 ) ( 4 )8 2x y xy
P Pxy B Pxy h y h yI wh wh
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Example Beam 1 #4Beam
The shear force over the free end as the integration of shear gstress
No resultant normal force across the sectionsAll sections including the built in end are free to distort All sections, including the built-in end, are free to distort.
Then, find displacement for plane stress problems
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Example Beam 1 #5Beam
1 ( ) xx x y
PxyE E EI u
1 ( )
y
xy y x
E E EIPxy
E E EI
x
y
uxv
( )
1
x yz E
y
z
ywz
1
xy xyG
xy
zu vy x
yz
v wz y w u
42
zx
w ux z
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Example Beam 1 #6Beam
2
( )Pxy u Pxy Px yu dx f y
2
( )2
( )2
x
y
u dx f yEI x EI EIPxy v Pxy Pxyv dy g xEI y EI EI
2 2
2
( 4 )8xy
EI y EI EIP u vh yGI y x
2 22 2( 4 ) ( ) ( )
8 2 2P Px y Pxyh y f y g xGI y EI x EI
2 2 2 ( )8 2 2Ph Py Px f y PyGI GI EI y
2
2 2 2 2
( )2
( ) ( )
g xEI x
Ph P P f P
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2 2 2 2( ) ( ) ( ) ( )8 2 2 2Ph Px g x Py f y Py F x G yGI EI x EI y GI
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Example Beam 1 #7Beam
2 2 2 2( ) ( )Ph Px g x Py f y Py
2
8 2 2 2
( ) ( )8
GI EI x EI y GI
G x F y G FPhGI
2
( ) ( )
As const, ( ) const and ( ) const.8
8y
Ph F y F G x GGI
GI
2 2
2
( ) ( )2 2Px g x g x PxG GEI x x EI
2
3
(
(
) ( )2
)
Pxg x G dxEI
Px G C
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( )6Pg x xE
1G x CI
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Example Beam 1 #8Beam
2 2 2 2( ) ( )Ph Px g x Py f y Py
2
( ) ( )8 2 2 2
( ) ( )
g y y yGI EI x EI y GI
G x F y G FPh
2
( ) ( )
As const, ( ) const and ( ) const.8
8G x F y G F
Ph F y F G x GGI
GI
2 22 2
8( ) ( )
2 2 2 2
GIPy f y Py f y Py PyF FEI y GI y GI EI
2
( ) (2Pyf yGI
2
3 3
)2Py F dyEI
P P
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3 3
26(
6) Py Py F y C
GI Iy
Ef
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Example Beam 1 #9Beam
2 2 3 3
2( )Px y Px y Py Pyu f y F y C
2
2 2 3
1
( )2 2 6 6
( )2 2 6
u f y F y CEI EI GI EI
Pxy Pxy Pxv g x G x CEI EI EI
2
2 2 6Boundary condition at the built-in end:
, 0, 0 0
EI EI EI
x L y u v C
3
1
2 3
6PLC G LEI
d PL PL 2 3
1
2
, 0, 0 , 2 3
dv PL PLx L y G Cdx EI EI
Ph 2 2 2Ph PL PhG F F G
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8G
8 2 8
G F F GI GI EI GI
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Example Beam 1 #10Beam
2 3 3 2 2
( )2 6 6 2 8Px y Py Py PL Phu yEI GI EI EI GI
2 3 2 3
2 6 6 2 8
Ans2 6 2 3
EI GI EI EI GIPxy Px PL x PLvEI EI EI EI
The relationship at 0 can be predicted by simple beam.These are displacements without shear of neutral plane
v y These are displacements without shear of neutral plane.
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Example Beam 1 #11Beam
2P Ph2 2
2
( 4 ) at NA8 8
Deflections of NA due to shear strain ( )
xy xyP Phh yGI GI
Ph L x
3 2 3 2
0
Deflections of NA due to shear strain ( )8
( ) Ans6 2 3 8y
L xGI
Px PL x PL Phv L xEI EI EI GI
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6 2 3 8y EI EI EI GI
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Example Beam 1 #12Beam
Let 5 m, 1 m, 0.3 m, 0.2 MN,
200 GPa 0 3
L h w PEE G
2 2
200 GPa, 0.3, 2(1 )
, 0, ( 4 )81x y xy
E G
Px Py h yI 81x y xyI
x
xy
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