Lecture Method of consistent deformation.pdf

13. Method of Consistent Deformations By James C. Maxwell in 1864

13. Method of Consistent Deformations

1

13.1 Structures with a Single Degree of Indeterminacy

32k A

C

B 10’

E = 30,000 ksi I = 512 in4

10’

1. Free-Body Diagram

MA

32 k

Ax Ay

Cy

Unknown variables = 4 (A x , A y , M A , C y ) Equations of Equilibrium = 3 (∑ F x = 0,∑ F y = 0, ∑ M = 0) →

Statically Indeterminate Structure

Degree of Indeterminacy = 4 -3 =1 = Number of Redundancy 13. Method of Consistent Deformations

2

Ma

32 k

Ax Cy

Ay 2. Selecting Cy as a redundant

MAO

Determinate Structure

32k

→

Axo

Primary Structure

∆ CO

Ayo Σ F x = 0

A xo = 0

Σ F y = 0

A yo = 32 k ↑

Σ M A = 0 M AO − 32 ×10 = 0 M AO = 320 k − kf


3

321 k-ft

32k ∆ CO

32k Bending Moment Diagram A

B

C

-320 Determining ∆CO using conjugate-beam method,

−

320 EI


4

−

∆ CO

=−

320 EI

=− ∆ CO

320 EI

1 2

2 3

×10 × × ( ×10 + 10)

26666.67 k − ft 3 EI

26666.67 × 123 = −3.0 in =− 30000 × 512

= 3.0 in ↓

3. Applying the redundant

MAC f CC

Axc

1 k Ayc


× C y

f CC : Flexibility Coefficient 5

Mac f CC

Axc

1k

× C y

Ayc Σ F x = 0

A xc = 0

Σ F y = 0

A yc = 1 k ↓

Σ M A = 0 − M AC + 1× 20 = 0 M AC = 20 k − kf

Bending Moment Diagram 10

20

× C y A

B


C

6

Determining f CC using conjugate-beam method,

20 EI f CC =

20 EI

1 2

2 3

× 20 × × ( × 20) 3

2666.67k − ft 3 = 2666.67 ×12 = 0.3in ↑ = 30000 × 512 EI ∆ CC = f CC × C y 4. Compatibility Condition

∆ C = ∆ CO + f CC C y = 0 ∆ CO = − f CC C y Unknowns : Forces Force Method Flexibility 13. Method of Consistent Deformations

Flexibility Method 7

C y = −

∆ CO f CC

=−

− 3.0

0.3

= 10k ↑

5. Reaction Forces

Using the equilibrium equations,

Σ F x = 0, A x = 0 Σ F y = 0, A y − 32 + 10 = 0, A y = 22 k ↑ Σ M A = 0, M A − 32 × 10 + 10 × 20 = 0, M A = 120 k − ft


8

Using the principle of superposition, 321 k-ft

32 k

3.0 in 32k 20 k-ft x 10

0.3 in ×10

1k x 10 MA = 120 k-ft

32 k

Ax = 0 Ay = 22 k 13. Method of Consistent Deformations

Cy = 10 k 9

6. Shear Force Diagram & Bending Moment Diagram

Using the reaction forces & the applied loads, Using the principle of superposition, A

B

C

-320

10 x 10

20 x 10 A

A -120 13. Method of Consistent Deformations

B

C

100

C

B 10

Moment as a redundant

Primary Structure°¡ Statically Determinate & StableÇ ´Ù ¸éÏ

, ¶²î¾

·Â Ý ¹

(È º ¤À

»·Â ³

)µ µ

Redundant°¡µÉ ¼öÀÖ À½

MA Ax Cy

Ay Selecting Ax as a redundant, MA

Cy

Ay Unstable Str uctur e


11

32k

32k

MA Ax

Cy

Ay Selecting MA as a redundant, 32k Axo Ayo

Cyo

Statically Deter mi nate & Stable Str uctur es


12

2. Selecting MA as a redundant

32k

θ AO

Axo Ayo Σ F x = 0

A xo = 0

Σ M A = 0

C yo = 16 k ↑

Σ F y = 0

A yo = 16 k ↑

θ AO

=−

PL2

16 EI

Cyo

= −0.0075 rad


13


1 k − ft AxA

f AA

AyA

Σ F x = 0

× M A CyA

A xA = 0

1 k ↓ Σ M A = 0 20 1 A k ↑ = Σ F y = 0 yA 20 C yA =

f AA

=

L

3 EI

= 0.0000625 rad / k − ft

θ AA = f AA × M A 13. Method of Consistent Deformations

14

4. Compatibility Condition

θ A = θ AO + f AA M A = 0 θ AO = − f AA M A M A = −

θ AO f AA

0.0075 = 120 k − ft = 0.0000625

5. Reaction Forces

Using the equilibrium equations, Using the principle of superposition, 6. Shear Force Diagram & Bending Moment Diagram

Using the reaction forces & the applied loads, Using the principle of superposition,


15

Example 13.1

M EI = constant

A L 1. Free-Body Diagram

B

M

MA Ax Ay

Cy

Unknown variables = 4 Equations of Equilibrium = 3 →


Degree of Indeterminacy = 4 -3 =1 13. Method of Consistent Deformations

16

M

MA Ax

By

Ay 2. Selecting By as a redundant

M

MAO

Determinate Structure →

Primary Structure

∆ BO

∆ BO =


17


MAB

f BB

1

× B y

AyB

f BB = 4. Compatibility Condition

∆ B = ∆ BO + f BB By = 0 B y = −∆ BO / f BB = 5. Reaction Forces 6. Shear Force Diagram & Bending Moment Diagram 13. Method of Consistent Deformations

18

Example 13.2

60 kN

15 kN/m

E = 200 GPa A

B

10 m

C

5m

D

I = 700×106 mm4

5m


60 kN

15 kN/m Ax Ay

By

Dy




19

60 kN

15 kN/m Ax

By

Ay

Dy

2. Selecting By as a redundant

60 kN

15 kN/m

∆ BO

Axo Ayo

Dyo

∆ BO =


20


f BB

AxB

1 AyB

× B y DyB

f BB = 4. Compatibility Condition

∆ B = ∆ BO + f BB By = 0 B y = −∆ BO / f BB = 5. Reaction Forces 6. Shear Force Diagram & Bending Moment Diagram 13. Method of Consistent Deformations

21

Example 13.3

E

F

28 k

15 ft A

B

C 25 k 3 @ 20 ft


D

25 k

22


28 k

Ax

Ay

25 k

25 k

Dy

Dx

Unknown variables = m + r = 13 Equations of Equilibrium = 2j = 12 →


Degree of Indeterminacy = 13 – 12 = 1


23

28 k

Ax

Ay

25 k

25 k

Dy

Dx

2. Selecting Dx as a redundant

28 k

∆ DO Axo 25 k Ayo


25 k Dyo

24

F O

∆ = ∑ F v

L AE

F

u D

∆ DO = ∑ u D

L AE

F O =


25


× D x f DD

u D

AxD

1 DyD

AyD

∆ = ∑ F v f DD =

∑

L AE

u D

2

F

L AE

=


26


∆ D = ∆ DO + f DD Dx = 0 D x = −∆ DO / f DD = 5. Reaction Forces 6. Shear Force Diagram & Bending Moment Diagram


27

Example 13.4

3 k/ft C

B

EI = constant

20 ft

A 30 ft


28


3 k/ft Cy

Ax

Cx

Ay



Degree of Indeterminacy = 4 -3 =1


29

2. Selecting Ax as a redundant

3 k/ft Cx Cy

∆ AO Ay 13. Method of Consistent Deformations

30

∆ = ∑ ∫ M v

M O

M EI

dx

m A

∆ AO = ∑ ∫ m A

M O EI

dx =


31


CxA m A

CyA

× Ax

1 f AA AyA

∆ = ∑ ∫ M v f AA =

M EI

dx

m A2

∑ ∫ EI dx =


32


∆ A = ∆ AO + f AA Ax = 0 A x = −∆ AO / f AA = 5. Reaction Forces 6. Shear Force Diagram & Bending Moment Diagram


33

13.2 Internal Forces and Moments as Redundants

10 k

10 k

A

12 k

EI = constant C

B 6 ft

8 ft

6 ft

20 ft

10 ft


10 k Ax Ay

10 k

12 k By

Cy




34

B

M B

θ BL = θ BR

θ BL

θ BR


θ Brel = θ BR − θ BL =0

35

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

10 k

10 k

12 k

Determinate Structure →

θ BOL

Primary Structure

θ BOR

θ BOL ≠ θ BOR

θ BOrel = θ BOR − θ BOL ≠ 0 θ BOL =

420 k − ft 2

θ BOrel = −

EI

θ BOR = −

533.33 k − ft 2 EI

953.33 k − ft 2 EI


36


1

1

× M B f BBL f BBR 2

6.67 k − ft / k − ft f = − BBR f BBL = EI

f BBrel = −

10 k − ft 2 / k − ft EI

16.67 k − ft 2 / k − ft EI


37


θ Brel = θ BOrel + f BBrel M B = 0 M B = −θ BOrel / f BBrel = −57.19 k − ft 5. Reaction Forces

- Equilibriu Equilibrium m Equations Equations of Members Members & Joints Joints 10

Ay = 7.14

10 57.19

ByAB = 12.86

57.19 57.19

12.86

5.91

57.19

ByBC = 5.91

12

Cy = 6.09

By = 18.77 6. Shear Force Diagram & Bending Moment Diagram 13. Method of Consistent Deformations

38

Internally Indeterminate Structures

Selecting Selecting a reaction reaction as a redundant redundant OR

Selecting an internal force or moment as a redundant

Externally Externally determinate, determinate, but Internally Internally indeterminate indeterminate structures structures Selecting Selecting a reaction reaction as as a redunda redundant nt (×) Selecting Selecting an internal internal force or moment moment as a redundant redundant D

P

C

A


B

39

1. Static Determinacy

Externally Static Determinacy

D

P

C

r=3 Statically Determinate Externally

Internally Static Determinacy

A

B

m = 6, r = 3, j = 4 m + r > 2j Statically Indeterminate


40

2. Selecting Internal Force FAD as a redundant

D

P

C

F O

A

B

∆ = ∑ F v

∆ ADO

D C

L AE

F

F ADO = 0 u AD , AD = 1.0

F O : Internal Forces due to External Forces

1

of the Primary System u AD : Internal Forces due to unit axial Forces of the Primary System

1

A

u AD

B


∆ ADO = ∑ u AD

L AE

F O = 41


D C

1

A

× F AD

1

u AD

B

∆ = ∑ F v

L AE

F

f AD, AD

f AD , AD =

∑u

2

AD

L AE

=


∆ AD = ∆ ADO + f AD , AD F AD = 0 F AD = −∆ ADO / f AD, AD = 13. Method of Consistent Deformations

42

5. Member Forces

F = F O + u AD F AD

6. Reaction Forces

R = RO + r AD F AD


43

Exampl e 13.5


44

Exampl e 13.6


45

13.3 Structures with Multiple Degrees of Indeterminacy

w A

B

C

D

E


w Ax Ay

By

Cy

Dy

Ey



Degree of Indeterminacy = 6 - 3 =3 13. Method of Consistent Deformations

46

w Ax Ay

By

Cy

Dy

Ey

2. Selecting By, Cy and Dy as the redundants

w AxO

∆ BO AyO

∆ CO

∆ DO

EyO

∆ BO = ∆ CO = ∆ DO =


47


f BB

AxB

f CB

f DB

1 AyB

f BB = f BC

AxC

f CB = f CC

f DB = f DC

1 AyC

f BC =

f BD

f CC = f CD

AxD

f DC = f DD

1 AyD

f BD =

f CD =


f DD =

× B y EyB

× C y EyC

× D y EyD

48


∆ BO + f BB B y + f BC C y + f BD Dy = 0 ∆ CO + f CB B y + f CC C y + f CD Dy = 0 ∆ DO + f DB B y + f DC C y + f DD Dy = 0 No. of Unknown Redundants = 3 No. of Compatibility Equations = 3 


By, Cy, Dy

49

5. Reaction Forces 6. Shear Force Diagram & Bending Moment Diagram Deflections to be calculated for determining unknown reducdants

∆ BO =

∆ CO =

∆ DO =

f BB =

f CB =

f DB =

f BC =

f CC =

f DC =

f BD =

f CD =

f DD =

12 13. Method of Consistent Deformations

50

According to the Maxwell’s Law of Reciprocal Deflections,

∆ BO =

∆ CO =

∆ DO =

f BB =

f CB =

f DB =

f BC =

f CC =

f DC =

f BD =

f CD =

f DD =

9 Homework 4 (10 Points)

Definition & Proof - Betti’s Law - Maxwell’s Law of Reciprocal Deflections 13. Method of Consistent Deformations

51

Exampl e 13.7


52

Exampl e 13.8


53

Exampl e 13.9


54

Example 13.10


55

Example 13.11


56

13.4 Support Settlements and Temperature Changes Support Settlements

w A

∆ B

B

∆ C

C

D

2. Selecting By and Cy as the redundants

w

∆ CO

∆ BO ∆ BO =

∆ CO =


57


f BB

f CB

1 f BB = f BC

f CB = f CC

1 f BC =

× B y

× C y

f CC =


∆ BO + f BB B y + f BC C y = ∆ B ∆ CO + f CB B y + f CC C y = ∆ C 13. Method of Consistent Deformations

58

A

∆ A

B

∆ B

D

C

∆D

∆ C C h o r d o f P r i m a y r B ea m

∆ BR

∆ CR

Rigid-Body Displacement

∆ BO + f BB B y + f BC C y = ∆ BR ∆ CO + f CB B y + f CC C y = ∆ CR Relati ve Displacement


59

Ri gid-Body Di splacement

A

B

 No

Bending Deformations

 No

Member Forces (Bending Moments)

 No

Stresses

C

D

Support Settlements in a Externally Determinate Structure  No

Effects


60

Example 13.12


61

Temperature Changes

Support Settlements or Temperture Changes in a Determinate Structure  No

Effects

Support Settlements or Temperture Changes in a Indeterminate Structure 

Resulting Stresses

Externally Indeterminate, and Internally Indeterminate 

Reactions

≠

0



Internal Forces

≠

0

Exter nally Deter minate, and I nter nally I ndeter minate  Reactions = 0  I nter nal F orces

0

Externally Determinate, and Internally Determinate 

Reactions = 0



Internal Forces = 0


62

Example 13.13

C

A

∆T 1

∆T 2

D

B

1. Static Determinacy Externally Static Determinacy

Internally Static Determinacy

r=3

m = 6, r = 3, j = 4

Statically Determinate Externally

m + r > 2j Statically Indeterminate


63

2. Selecting Internal Force FAD as a redundant

∆T 1

C

∆ ADO

∆T 2

A

C

∆ = ∑ F vα ( ∆T ) L

D

B

1

D

∆ ADO = ∑ u ADα (∆T ) L = 1

A

u AD

B


64


C

1

1

A

u AD

D

B

∆ = ∑ F v

× F AD

L AE

F

f AD, AD

f AD , AD =

∑

2 u AD

L AE

=


∆ AD = ∆ ADO + f AD, AD F AD = 0 F AD = −∆ ADO / f AD, AD = 13. Method of Consistent Deformations

65

5. Member Forces

F = F O + u AD F AD = u AD F AD 6. Reaction Forces

R = RO + r AD F AD = r AD F AD = 0


66

13.5 Method of Least Work

w Ax

A Ay

B

C

By

Cy

Selecting By as the redundant

w

By w

By 13. Method of Consistent Deformations

67

U = f ( w)

U = f ( w, By )

According to the Castigliano’s Second Theorem,

∂U =0 ∂ B y



Compatibility Condition

 F or the value of the r edundant that satisf ies the equi librium equation and compatibil ity,

the str ain ener gy of the str uctur e is a maxi mum or mini mum.  MINIMUM  Th e magni tude of the r edundants of a statically in deter minate str uctur e

must be such that the str ain ener gy (i nter nal work) is a mini mum (the least).  Principle of L east Work


68

For structures with n redundants, U = f ( w, R1 , R2 , R3 ,L, Rn )

∂U =0 ∂ R1

∂U =0 ∂ R2

∂U =0 ∂ R3


L

∂U =0 ∂ Rn

69

Example 13.14

30 kN/m A 10 m

80 kN B 5m

EI = constant D

C 5m

Free-Body Diagram

30 kN/m Ax Ay

80 kN By

Dy



Degree of Indeterminacy = 4 - 3 = 1 13. Method of Consistent Deformations

70

Selecting By as the redundant

30 kN/m Ax

80 kN By Dy

Ay A x = 0

U =

A y = 245 − 0.5B y

L

D y = 135 − 0.5B y

M 2

∫0 2 EI dx

∂U = ∂ B y

L

∫0

∂ M M dx = 0 ∂ B y EI


71

30 kN/m

80 kN B

A

By

D

C

135 – 0.5By

245 – 0.5By

X coordinates

Segment

Origin Limit (m)

AB

A

DC CB

D

∂U = ∂ B y

D L

∫0

∂ M / ∂ B y

M

(245 − 0.5 B y ) x − 15 x 2 (135 − 0.5 B y ) x 0−5 5 − 10 (135 − 0.5 B y ) x − 80( x − 5)

0 − 10

∂ M M dx = ∂ B y EI

10

5

10

∫0 Ldx + ∫0 Ldx + ∫5 Ldx

− 0.5 x − 0.5 x

− 0.5 x

=0

B y = 242.5 kN (↑) A y =

D y =


72

Example 13.15


73

Example 13.16


74

Homework 5 (10 Points)

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


75

Lecture Method of consistent deformation.pdf

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