Lecture Method of consistent deformation.pdf

13. Method of Consistent Deformations 1

13. Method of Consistent Deformations

By James C. Maxwell in 1864


13.1 Structures with a Single Degree of Indeterminacy

AB

C

32kE = 30,000 ksi

I = 512 in4

10 10

1. Free-Body Diagram

32 k

AxAy Cy

MA

)yAyx C,M,A,(A

)0,0,0( === MFF yxUnknown variables = 4

3Equations of Equilibrium =

Statically Indeterminate Structure

Degree of Indeterminacy = 4 -3 =1

= Number of Redundancy


32 k

Ax

Ay Cy

Ma

2. Selecting Cy as a redundant

32k

Axo

Ayo

MAO Determinate Structure

Primary Structure

CO

0= xF 0=xoA= kAyo 320 y =F

kfkM AO = 32001032 =AOM0= AM


32k

32k

321 k-ft

CO

Bending Moment Diagram

A B C

-320

Determining CO using conjugate-beam method,

EI320


EI320

)101032

2110320 += (

EICO

EI67.26666=

512300001267.26666 3

=

3ftk in0.3=

= in.CO 033. Applying the redundant

Axc

MACCCf

yC1 k

Ayc fCC : Flexibility Coefficient


Axc

Ayc

MacCCf

1kyC

0=xcA0= xF= kAyc 10 y =F

kfkM AC = 200201 =+ ACM0= AM

Bending Moment Diagram

A B C

1020

yC


Determining fCC using conjugate-beam method,

EI20

)2032(

212020 =

EIfCC

EIftk 367.2666 = 51230000

1267.2666 3

= = in3.0

yCCCC Cf =

4. Compatibility Condition

0=+= yCCCOC CfyCCCO Cf=

Unknowns : Forces Force Method

Flexibility MethodFlexibility


=== kf

CCC

COy 103.0

0.3

5. Reaction Forces

Using the equilibrium equations,

0A ,0 == xxF==+= k22A0,1032 ,0 yyy AF

120 ,020101032 ,0 ftkMMM AAA ==+=


Using the principle of superposition,

32 k

32k

321 k-ft

in0.3

1k x 10

20 k-ft x 10103.0 in

32 k

Ay = 22 k

MA = 120 k-ft

Ax = 0

Cy = 10 k


6. Shear Force Diagram & Bending Moment Diagram

Using the reaction forces & the applied loads,

Using the principle of superposition,A B C

-320

A B C

20 x 1010 x 10

A

B

C

-120

100


Moment as a redundantPrimary Structure Statically Determinate & Stable, ()Redundant

AxAy Cy

MA

Selecting Ax as a redundant,

Ay Cy

MA

Unstable Structure


AxAy Cy

MA

32k

32k

Selecting MA as a redundant,

Ayo Cyo

32k

Axo

Statically Determinate & Stable Structures


2. Selecting MA as a redundant

Axo

Ayo Cyo

32k

AO

0=xoA0= xF= kCyo 160= AM= kAyo 160 y =F

EIPL

16

2

= rad0075.0=AO


3. Applying the redundant

AxA

AyACyA

ftk 1AAf

AM

0=xAA0= xF= kCyA 20

10= AM

= kAyA 201

0 y =F

EIL

3=AAf ftkrad = /0000625.0

AAAAA Mf =



0=+= AAAAOA MfAAAAO Mf=

0000625.00075.0=

AA

AOA f

M = ftk =120

5. Reaction ForcesUsing the equilibrium equations,



Using the reaction forces & the applied loads,



Example 13.1

AB

L

M

EI = constant


AxAy Cy

MAM

Unknown variables = 43Equations of Equilibrium =




AxAy By

MAM

2. Selecting By as a redundantMMAO

BO

Determinate Structure

Primary Structure

=BO



MABBBf

yB1

AyB

=BBf4. Compatibility Condition

0=+= yBBBOB Bf== BBBOy fB /

5. Reaction Forces



Example 13.2

15 kN/m60 kN

AB C

D

10 m 5 m 5 m

E = 200 GPa

I = 700106 mm4


AxAy

By

15 kN/m60 kN

Dy





AxAy

By

15 kN/m60 kN

Dy

2. Selecting By as a redundant

15 kN/m60 kN

AxoAyo Dyo

BO

=BO



1yBAxB

AyB DyB

BBf

=BBf4. Compatibility Condition

0=+= yBBBOB Bf== BBBOy fB /

5. Reaction Forces



Example 13.3

25 k25 k

28 k

3 @ 20 ft

15 ft

A B C D

E F



25 k25 k

28 k

AxAy Dy

Dx

m + r Unknown variables = = 13

Equations of Equilibrium = 2j = 12


Degree of Indeterminacy = 13 12 = 1


25 k25 k

28 k

AxAy Dy

Dx

2. Selecting Dx as a redundant

25 k25 k

28 k

Axo

Ayo Dyo

DO


= FAELFvOF

Du

== ODDO FAELu



1AxD

AyD DyD

DDf

xD

= FAELFv

Du

== AELuf DDD 2


4. Compatibility Condition0=+= xDDDOD Df

== DDDOx fD /

5. Reaction Forces



Example 13.4

3 k/ft

30 ftA

B CEI = constant

20 ft


1. Free-Body Diagram3 k/ft

AxAy

CyCx





2. Selecting Ax as a redundant3 k/ft

Ay

Cy

Cx

AO


= dxEIMM vOM

Am

== dxEIMm OAAO



1

AyA

CyA

CxA

AAf

= dxEIMM v

Am xA

== dxEImf AAA2


4. Compatibility Condition0=+= xAAAOA Af

== AAAOx fA /

5. Reaction Forces



13.2 Internal Forces and Moments as Redundants

10 k

A B C

10 k 12 k

6 ft 8 ft 6 ft 20 ft 10 ft

EI = constant


10 k 10 k 12 k

AyBy Cy

Ax





BM

B

BRBL =BL

BR Brel BLBR = 0=


2. Selecting Internal Moment (Bending Moment) MB as a redundant

10 k 10 k 12 k

BOL BOR

Determinate Structure

Primary Structure

BORBOL BOrel BOLBOR = 0

EIftk

BOR

233.533 =EI

ftkBOL

2420 =

EIftk

BOrel

233.953 =


3. Applying the redundant1 1

BBLf BBRf

BM

EIftkftk

fBBR= /10

2

EIftkftk

fBBL= /67.6

2

EIftkftk

fBBrel= /67.16

2


4. Compatibility Condition0=+= BBBrelBOrelBrel Mf

ftkfM BBrelBOrelB == 19.57/5. Reaction Forces

- Equilibrium Equations of Members & Joints

57.19 57.1910 10 57.19 1257.19

Ay = 7.14 ByAB = 12.86 Cy = 6.09By

BC = 5.9112.86 5.91

By = 18.77



Internally Indeterminate StructuresSelecting a reaction as a redundant

OR Selecting an internal force or moment as a redundant

Externally determinate, but Internally indeterminate structures

Selecting a reaction as a redundant ()

Selecting an internal force or moment as a redundant

A B

CD

P


1. Static Determinacy

Externally Static Determinacy

r = 3

Statically Determinate Externally

A B

CD

P

Internally Static Determinacy

m = 6, r = 3, j = 4

m + r > 2j

Statically Indeterminate


2. Selecting Internal Force FAD as a redundant

= FAELFv

== OADADO FAELu

A B

CD

P

ADOOF

A B

CD

1ADu

1

FO : Internal Forces due to External Forces

of the Primary System

uAD : Internal Forces due to unit axial Forces

of the Primary System

ADOF 0=ADADu , 0.1=



A B

CD

1

1

ADADf ,

= FAELFv

ADu

ADF

== AELuf ADADAD 2,4. Compatibility Condition

0, =+= ADADADADOAD Ff== ADADADOAD fF ,/


5. Member Forces

ADADO FuFF +=

6. Reaction Forces

ADADO FrRR +=


Example 13.5


Example 13.6


13.3 Structures with Multiple Degrees of Indeterminacy

w

A B C D E


AxAy

By Cy

w

Dy Ey



Degree of Indeterminacy = 6 - 3 =3


AxAy

By Cy

w

Dy Ey

2. Selecting By, Cy and Dy as the redundants

w

AxOAyO EyO

BO CO DO

=BO=CO=DO



1yBBBfAxB

AyBEyB

CBf DBf

=BBf =CBf =DBf

1yCBCfAxC

AyCEyC

CCf DCf

=BCf =CCf =DCf

1yDBDfAxD

AyDEyD

CDf DDf

=BDf =CDf =DDf



0=+++ yBDyBCyBBBO DfCfBf0=+++ yCDyCCyCBCO DfCfBf0=+++ yDDyDCyDBDO DfCfBf

No. of Unknown Redundants = 3

No. of Compatibility Equations = 3

By, Cy, Dy


5. Reaction Forces


Deflections to be calculated for determining unknown reducdants

=DO=CO=BO=DBf=CBf=BBf=DCf=BCf =CCf

=BDf =CDf =DDf

12


According to the Maxwells Law of Reciprocal Deflections,

=BO =CO =DO=BBf =CBf =DBf=BCf =CCf =DCf=BDf =CDf =DDf

9Homework 4 (10 Points)Definition & Proof

- Bettis Law

- Maxwells Law of Reciprocal Deflections


Example 13.7


Example 13.8


Example 13.9


Example 13.10


Example 13.11


13.4 Support Settlements and Temperature Changes

Support Settlementsw

A B C DB C

2. Selecting By and Cy as the redundantsw

BO CO

=CO=BO



1

BBf CBfyB

=CBf=BBf

1

BCf CCf

=BCf =CCfyC


=++ yBCyBBBO CfBf B=++ yCCyCBCO CfBf C


A B C D

AB

C D

=++ yBCyBBBO CfBf=++ yCCyCBCO CfBf

Rigid-Body Displacement

Chord of Primary BeamBR

CR

BRCR

Relative Displacement


Rigid-Body Displacement

A B C D

No Bending Deformations No Member Forces (Bending Moments) No Stresses

Support Settlements in a Externally Determinate Structure No Effects


Example 13.12


Temperature Changes

Support Settlements or Temperture Changes in a Determinate Structure

No EffectsSupport Settlements or Temperture Changes in a Indeterminate Structure

Resulting Stresses

Externally Indeterminate, and Internally Indeterminate

Reactions 0 Internal Forces 0

Externally Determinate, and Internally Indeterminate Reactions = 0 Internal Forces 0

Externally Determinate, and Internally Determinate

Reactions = 0 Internal Forces = 0


Example 13.13

A B

C D1T

2T

1. Static Determinacy

Externally Static Determinacy

r = 3

Statically Determinate Externally

Internally Static Determinacy

m = 6, r = 3, j = 4

m + r > 2j

Statically Indeterminate


2. Selecting Internal Force FAD as a redundant

A B

C D1T

2TADO

= LTFv )(

A B

C D

1

1

ADu

== LTuADADO )(



A B

C D

1

1ADF

ADADf ,

= FAELFv

ADu

== AELuf ADADAD 2,4. Compatibility Condition

0, =+= ADADADADOAD Ff== ADADADOAD fF ,/


5. Member Forces

ADADFu=ADADO FuFF +=

6. Reaction Forces

ADADFr=ADADO FrRR += 0=


13.5 Method of Least Workw

A B CAxAy By Cy

Selecting By as the redundantw

By

w

By


=U ),( yBwf)(wf=UAccording to the Castiglianos Second Theorem,

=

yBU

0 Compatibility Condition

For the value of the redundant that satisfies the equilibrium equation and compatibility,the strain energy of the structure is a maximum or minimum.

MINIMUM The magnitude of the redundants of a statically indeterminate structure

must be such that the strain energy (internal work) is a minimum (the least).

Principle of Least Work


For structures with n redundants,

),,,,,( 321 nRRRRwf L=U0

1

=RU 0=

nR

U02

=RU 0

3

=RU L


Example 13.1430 kN/m 80 kN

10 m 5 m 5 mB C

EI = constant

A D

AxAy

By Dy

Free-Body Diagram30 kN/m 80 kN



Degree of Indeterminacy = 4 - 3 = 1


Selecting By as the redundant

30 kN/m 80 kN

Ax

Ay Dy

By

yy BD 5.0135=0=xA yy BA 5.0245=

L dxEIM02

2=U

L

y

dxEIM

BM

00==

yB

U


30 kN/m 80 kN

245 0.5By 135 0.5By

ByA

B

C D

(m)Limit

scoordinate XSegment

215)5.0245( xxBy xBy )5.0135(

)5(80)5.0135( xxBy

yBM /MOriginx5.0100AABx5.0D 50 DCx5.0105CB D

=L

y

dxEIM

BM

0=

yB

U 100 dxL + 50 dxL + 105 dxL 0=)(5.242 = kNBy

=yA =yD


Example 13.15


Example 13.16


Homework 5 (10 Points)

Why cannot the method of least work be used for analyzing the effects of support settlements and temperature changes?


Homework 6 (40 Points)Solve the following problems.

13.1 ~ 13.12 : 2 problems13.13 ~ 13.25 : 2 problems13.26 ~ 13.36 : 2 problems13.37 ~ 13.45 : 2 problems13.46 ~ 13.47 : 1 problem13.48 : 1 problem13.49 ~ 13.61 : 2 problems

12 problems

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Lecture Method of consistent deformation.pdf