3
1 Chapter 4: Deflection and Stiffness I am never content until I have constructed a mechanical model of the subject I am studying. If I succeed in making one, I understand; otherwise I do not. William Thomson (Lord Kelvin) Spring Rate Prismatic Bars - Analogy to Coil Spring The force and displacement Relation: Torsional Spring L AE k L AE P AE PL / where = = = δ δ δ P P P = k δ L GJ k L GJ T GJ TL t / where = = = θ θ Deflection due to Bending ( ) [ ] 2 2 2 2 2 1 1 dx y d dx dy dx y d EI M + = = ρ () x f y dx dy dx y d EI M dx y d EI V dx y d EI q = = = = = θ 2 2 3 3 4 4 Boundary Conditions BCs ( ) ( ) ( ) 0 0 0 0 0 = = = = = = x dx x dy x y θ BCs ( ) ( ) ( ) ( ) 0 0 3 3 2 2 = = = = = = = = dx L x y d L x V dx L x y d L x M BCs BCs ( ) 0 0 = = x y ( ) ( ) 0 0 0 2 2 = = = = dx x y d x M ( ) ( ) 0 2 2 = = = = dx L x y d L x M ( ) 0 = = L x y M=0 Essential Boundary Conditions

Deflection due to Bending Boundary Conditionspkwon/me471/Lect 4.1.pdf2! Superposition: Cantilever Beam! P w M Slope Maximum Deflection Elastic Curve EI PL 2 2 max − θ = EI wL 6

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Page 1: Deflection due to Bending Boundary Conditionspkwon/me471/Lect 4.1.pdf2! Superposition: Cantilever Beam! P w M Slope Maximum Deflection Elastic Curve EI PL 2 2 max − θ = EI wL 6

1

Chapter 4: Deflection and Stiffness

I am never content until I have constructed a mechanical model of the subject I am studying. If I succeed in making one, I understand; otherwise I do not. William Thomson (Lord Kelvin)

Spring Rate •  Prismatic Bars - Analogy to Coil Spring

–  The force and displacement Relation:

•  Torsional Spring

LAEkLAEP

AEPL

/where =

== δδ

δ

P

P P = k δ

LGJkLGJT

GJTL

t /where =

== θθ

Deflection due to Bending

( )[ ] 2

2

2

22

11

dxyd

dxdydxyd

EIM

≈+

==ρ

( )xfydxdydxyd

EIM

dxyd

EIV

dxyd

EIq

=

=

=

=

=

θ

2

2

3

3

4

4

Boundary Conditions

BCs

( )( ) ( ) 000

00

====

==

xdxxdy

xy

θ

BCs

( ) ( )

( ) ( ) 0

0

3

3

2

2

==

==

==

==

dxLxydLxV

dxLxydLxM

BCs

BCs ( ) 00 ==xy

( ) ( ) 000 2

2=

===

dxxydxM ( ) ( ) 02

2=

===

dxLxydLxM

( ) 0== Lxy

M=0 Essential Boundary Conditions

Page 2: Deflection due to Bending Boundary Conditionspkwon/me471/Lect 4.1.pdf2! Superposition: Cantilever Beam! P w M Slope Maximum Deflection Elastic Curve EI PL 2 2 max − θ = EI wL 6

2

Superposition: Cantilever Beam

P

w

M

Slope Maximum Deflection Elastic Curve

EIPL2

2

max−

EIwL6

3

max−

EIML2

max −=θ

EIPLy3

3

max−

=

EIwLy8

4

max−

=

EIMLy2

2

max −=

( )23 36

LxxEIPy −=

( )222

6424

LLxxEIwxy +−

−=

EIMxy2

2

−=

Superposition: Simple Supported Beam

P

w

M

Slope Maximum Deflection Elastic Curve

EIPL16

2

max−

EIwL24

3

max−

EIMLEIML

A

B

6

3

=

−=

θ

θ

EIPLy48

3

max−

=

EIwLy

3845 4

max−

=

EIMLy243

2

max−

=

( )xLxEIPy 23 3448

−=

( )xLLxxEIwy 334 2

24+−

−=

( )xLxEILMy 23

6−

−=

A B

2/Lx <

4.7 Strain Energy

•  Work of a Force

•  Work of a Couple Moments

•  Strain Energy

Δ== ∫ PFdxUx

e 21

0

θθ MMdUx

e 21

0== ∫

Normal Stress

Shear Stress

Multi-axial Stress

dVE

dVUVVi ∫∫ ==22

1 2σσε

dVG

dVUVVi ∫∫ ==22

1 2ττγ

[ ]

( ) ( ) ( ) dVGE

vE

dVU

V xzyzxyzxyzyxzyx

V yzyzxzxzxyxyzzyyxxi

⎥⎦

⎤⎢⎣

⎡ +++++−++=

+++++=

222222

21

21

21

τττσσσσσσσσσ

γτγτγτεσεσεσ

Important - Relate to Fracture Toughness

4.7 Elastic Strain Energy •  Axial Load

barprismaticafor2222

2

0

2

2

22

AELNdx

AENdV

EANdV

EU

L

VVx

i ==== ∫∫∫σ

•  Bending Moment

dxEIMdAydx

EIMdAdx

IMy

EdVE

ULL

AVVx

i ∫∫ ∫∫∫ ==⎟⎠

⎞⎜⎝

⎛==0

2

0

22

222

2221

•  Transverse Shear

•  Torsion Moment dx

GJTdAdx

GJTdAdx

JT

GdV

GU

LL

AVVi ∫∫ ∫∫∫ ==⎟⎠

⎞⎜⎝

⎛==0

2

0

22

222

2221

ρτ

dxGACVdA

tQdx

GIVdAdx

ItVQ

GdV

GU

LL

AVVi ∫∫ ∫∫∫ =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛==0

2

0

2

2

222

2221

Page 3: Deflection due to Bending Boundary Conditionspkwon/me471/Lect 4.1.pdf2! Superposition: Cantilever Beam! P w M Slope Maximum Deflection Elastic Curve EI PL 2 2 max − θ = EI wL 6

3

Castigliano’s Theorem

Castigliano Theorem

bending1

torsion1

ncompressioandtension1

dxFMM

EIFU

dxMTT

GJMU

dxFFF

AEFU

iii

iii

iii

⎟⎟⎠

⎞⎜⎜⎝

∂=

∂=

⎟⎟⎠

⎞⎜⎜⎝

∂=

∂=

⎟⎟⎠

⎞⎜⎜⎝

∂=

∂=

δ

θ

δ

Fictitious Force and Moment

4-10 Statically Indeterminate Problems

•  The law of statics alone cannot resolve a certain problem.

•  Use a compatibility equation ex. δδδ == 21