Class Notes on Structural Analysis-2 for Stiffness & Flexibility Method of Matrix By HSL

Class Note for Structural Analysis 2 Fall Semester, 2013

Hae Sung Lee, Professor Dept. of Civil and Environmental Engineering Seoul National University Seoul, Korea

Contents Chapter 1 Slope Deflection Method

1

1.0 Comparison of Flexibility Method and Stiffness Method………………………… 2 1.1 Analysis of Fundamental System………………………………………………..... 5 1.2 Analysis of Beams………………………………………………………………… 8 1.3 Analysis of Frames………………………………………………………………... 17 Chapter 2 Iterative Solution Method & Moment Distribution Method 2.1 Solution Method for Linear Algebraic Equations………………………………… 2.2 Moment Distribution Method……………………………………………………... 2.3 Example - MDM for a 4-span Continuous Beam…………………………………. 2.4 Direct Solution Scheme by Partitioning…………………………………………... 2.5 Moment Distribution Method for Frames………………………………………… Chapter 3 Energy Principles

32 33 37 42 44 45 47

3.1 Spring-Force Systems……………………………………………………………... 48 3.2 Beam Problems……………………………………………………………………. 49 3.3 Truss problems……………………………………………………………………. 53 Chapter 4 Matrix Structural Analysis 4.1 Truss Problems…………………………………………………………………… 4.2 Beam Problems…………………………………………………………………… 4.3 Frame Problems…………………………………………………………………... Chapter 5 Buckling of Structures 5.0 Stability of Structures……………………………………………………………... 5.1 Governing Equation for a Beam with Axial Force………………………………... 5.2 Homogeneous Solutions…………………………………………………………... 5.3 Homogeneous and Particular solution…………………………………………….. 5.4 Energy Method……………………………………………………………………. 5.5 Approximation with the Homogeneous Beam Solutions…………………………. 5.6 Nonlinear Analysis of Truss……………………………………………………….

56 57 68 76 79 80 81 82 85 86 89 91

School of Civil, Urban & Geosystem Eng., SNU

1

Chapter 1 Slope Deflection Method

A

B

Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

2

School of Civil, Urban & Geosystem Eng., SNU 2.0 Comparison of Flexibility Method and Stiffness Method Flexibility Method

Stiffness Method P

k1 z

P k2

k1

Remove redundancy (Equilibrium)

z

k2

Compatibility

P P

X δ1 = δ 2 = δ

z

z

Compatibility

Equilibrium

δ1 = δ 2

k1 X P− X = →X = P k1 + k 2 k1 k2

k1δ + k2δ = P → δ =

P k1 + k 2


3

School of Civil, Urban & Geosystem Eng., SNU Flexibility Method

L/2

L/2

P B

A

z

Stiffness Method

EI

EI

L

L z

Remove redundancy (Equilibrium)

B

A

C

P EI

EI

L

L

C

Compatibility

3PL 16

+

θB 1

+

θ BA = θ BC = θ B z

z

Compatibility δB0 =

2

L 1 PL PL 2L (1 + ) ×1 = , δ BB = 6 EI 2 4 16 EI 3EI

δ B 0 + δ BB M B = 0 → M B = −

δB0 3PL =− δ BB 32

Equilibrium

3EI 3PL f B B , M BC = 0 , M BA = M BC = θB L 16 ∑ M B =M BAf + M BCf + M BAB + M BCB = 0 →

f M BA =−

−

3PL 6 EI PL2 + θB = 0 → θB = 16 L 32 EI


4

School of Civil, Urban & Geosystem Eng., SNU Flexibility Method

Stiffness Method

1. Release all redundancies.

1.

Fix all Degrees of Freedom.

2. Calculate displacements induced by external loads at the released redundancies.

2.

Calculate fixed end forces induced by external loads at the fixed DOF.

3. Apply unit loads and calculate displacements at the released redundancies.

3.

Apply unit displacements and calculate member end forces at the DOFs.

4. Construct the flexibility equation by superposing the displacement based on the compatibility conditions.

4.

Construct the stiffness equation by superposing the member end forces based on the equilibrium equations.

5. Solve the flexibility equation.

5.

Solve the stiffness equation.

6. Calculate reactions and other quantities as needed.

6.

Calculate reactions and other quantities as needed.


5

School of Civil, Urban & Geosystem Eng., SNU 2.1 Analysis of Fundamental System 2.1.1 End Rotation

A

B

θA z

Flexibility Method i) θ B = 0 MA

MB

L L MA + M B = −θ A 3EI 6 EI → M A = − 4 EI θ A , M B = 2 EI θ A L L L L MA + MB = 0 6 EI 3EI ii) θ A = 0 MA = −

2 EI 4 EI θB , M B = θB L L

z

Sign Convention for M :Counterclockwise “+”

z

θ A ≠ 0 , θB ≠ 0

4 EI 2 EI θA + θB L L 2 EI 4 EI MB = θA + θB L L MA =

2.1.2 Relative motion of joints

∆


6

School of Civil, Urban & Geosystem Eng., SNU z

Flexibility Method

∆

L L ∆ MA + MB = − 3EI 6 EI L L L ∆ MA + MB = 6 EI 3EI L

z

=−

6 EI ∆ 6 EI ∆ , MB = L L L L

or in the new sign convention : M A =

6 EI ∆ 6 EI ∆ , MB = L L L L

→M

A

Final Slope-Deflection Equation 4 EI 2 EI 6 EI θA + θB + L L L 2 EI 4 EI 6 EI MB = θA + θB + L L L

MA =

z

∆ L ∆ L

In Case an One End is Hinged 4 EI 2 EI 6 EI θA + θB + L L L 2 EI 4 EI 6 EI MB = θA + θB + L L L

MA =

EI ∆ 2 EI 3EI ∆ =0→ θ A = − θB − L L L L L ∆ 3EI 3EI ∆ = θB + L L L L

2.1.3 Fixed End Force z

Both Ends Fixed


7


One End Hinged

MB

MA +

MA/2

MA 3 MA 2 z

3 MA 2

Ex.: Uniform load case with a hinged left end

M Bf = −

3qL2 qL2 qL2 qL2 − =− =− , 12 24 24 8

M Af = 0

2.1.4 Joint Equilibrium

Ffixed Fjoint

Fmember

Joint i

− ∑ F fixed −∑ Fmember + ∑ F joint = 0

or

∑F

fixed

+ ∑ Fmember = ∑ F joint


8

School of Civil, Urban & Geosystem Eng., SNU 2.2 Analysis of Beams 2.2.1 A Fixed-fixed End Beam

P B

A

C EI

a z

DOF : θ B , ∆B

z

Analysis

b

i) All fixed : No fixed end forces ii) θ B ≠ 0 , ∆ B = 0

2 EI 4 EI 4 EI 2 EI 1 1 θ B , M BA = θ B , M 1BC = θ B , M CB = θB a a b b 6 EI 6 EI 1 1 1 = −VAB = 2 θ B , − VBC = VCB = 2 θB a b

M 1AB = 1 VBA

2 EI θB a

4 EI θB a

6 EI θB a2

6 EI θB a2

4 EI θB b

6 EI θB b2

2 EI θB b

6 EI θB b2

iii) θ B = 0 , ∆ B ≠ 0

6 EI 6 EI 6 EI 6 EI 2 2 2 = 2 ∆ B , M BC = − 2 ∆B ∆ B , M BA = − 2 ∆ B M CB 2 a b a b 12 EI 12 EI 2 2 2 = −VAB = 3 ∆ B , VBC = −VCB = 2 ∆B a b

2 M AB = 2 VBA

6 EI ∆B a2

12 EI ∆B a3

6 EI ∆B a2

12 EI ∆B a3

6 EI ∆B b2

12 EI ∆B b3

6 EI ∆B b2

12 EI ∆B b3


9


Construct the Stiffness Equation

1 1 1 1 1 1 2 2 = 0 → M BA + M BC + M BA + M BC = 0 → 4 EI ( + )θ B + 6 EI ( 2 − 2 )∆ B = 0 a b a b 1 1 1 1 6 EI ( 2 − 2 )θ B + 12 EI ( 3 + 3 )∆ B = P ∑VBi = P → VBA1 + VBC1 + VBA2 + VBC2 = P → a b a b 2 2 3 3 (b − a )a b ab θB = − P , ∆B = P 3 2 EIl 3EIl 3

∑M

i B

ab 2 2 EI 6 EI θB + 2 ∆ B = 2 P , a a l a 2b 2 EI 6 EI = θB − 2 ∆ B = − 2 P b b l

2 M AB = M 1AB + M AB = 1 2 M CB = M CB + M CB

2.2.2 Analysis of a Two-span Continuous Beam (Approach I)

q

qL B

A

2EI

EI L

z

DOF : θ B , θC

z

Analysis

C

L

i) Fix all DOFs and Calculate FEM. f = M AB

qL2 qL2 qL2 qL2 f f f , M BA =− , M BC = , M CB =− 8 12 12 8

ii) θ B ≠ 0 , θC = 0 M 1AB =

2 EI 4 EI 8 EI 4 EI 1 1 1 θ B , M BA = θ B , M BC = θ B , M CB = θB L L L L

iii) θ B = 0 , θC ≠ 0 2 M BC =

z

4 EI 8 EI 2 θC , M CB = θC L L


∑M

i B

f f 1 1 2 = 0 → M BA + M BC + M BA + M BC + M BC =0→

qL2 EI EI + 12 θ B + 4 θ C = 0 L L 24

∑M

i C

f 1 2 = 0 → M CB + M CB + M CB =0→

−

qL2 EI EI + 4 θ B + 8 θC = 0 L L 8L


10

School of Civil, Urban & Geosystem Eng., SNU θB = − z

Member End Forces f M AB = M AB + M 1AB =

qL2 2 EI 3 + θ B = qL2 12 L 48

1 f M BA = M BA + M BA =−

qL2 4 EI 1 + θ B = − qL2 12 L 8

1 2 f M BC = M BC + M BC + M BC =

qL2 8 EI 4 EI 1 + θB + θC = qL2 8 L L 8

1 2 f M CB = M CB + M CB + M CB =−

z

qL3 qL3 , θC = 96 EI 48EI

qL2 4 EI 8 EI + θB + θC = 0 L L 8

Various Diagram

- Freebody Diagram

1 2 qL 16

1 2 qL 8

5 qL 8

9 qL 16

7 qL 16

3 qL 8

19 qL 16 - Moment Diagram

1 2 qL 16

17 2 qL 512

1 2 qL 8

3 2 qL 16

7 L 16


11

School of Civil, Urban & Geosystem Eng., SNU 2.2.3 Analysis of a Two-span Continuous Beam (Approach II)

q

qL B

A EI L z

DOF : θ B

z

Analysis

2EI

C

L

i) Fix all DOFs and Calculate FEM. f = M AB

qL2 qL2 qL2 1 qL2 3qL2 f f , M BA =− , M BC = + = 12 12 8 2 8 16

ii) θ B ≠ 0 M 1AB = z

Construct Stiffness Equation

∑M z

2 EI 4 EI 6 EI 1 1 θ B , M BA = θ B , M BC = θB L L L

B

1 1 f f = 0 → M BA + M BC + M BA + M BC =0

qL2 3qL2 EI EI qL2 + + 4 θB + 6 θB = 0 → θB = − 12 16 L L 96 EI

Member End Forces

qL2 2 EI 3 + θ B = qL2 12 L 48 2 qL 4 EI 1 1 f M BA = M BA + M BA =− + θ B = − qL2 12 L 8 2 3qL 6 EI 1 1 f M BC = M BC + M BC = + θ B = qL2 16 L 8 f M AB = M AB + M 1AB =


12

School of Civil, Urban & Geosystem Eng., SNU 2.2.4 Analysis of a Beam with an Internal Hinge (4 DOFs System) q

C

B

A EI l z

DOF : θ B , θCL , θCR , ∆C

z

Analysis

D EI

EI

l

l

i) All fixed M

f AB

ql 2 ql 2 f = , M BA = − 12 12

ii) θ B ≠ 0

M 1AB =

2 EI 4 EI 2 EI 6 EI 1 1 1 1 θ B , M BA = M BC = θ B , M CB = θ B , VCB = 2 θB l l l l

iii) θCL ≠ 0

2 M BC =

2 EI L 4 EI L 6 EI 2 2 θC , M CB = θC , VCB = 2 θCL l l l

iv) θCR ≠ 0

3 M CD =

4 EI R 2 EI R 6 EI 3 3 θC , M DC = θC , VCD = − 2 θCL l l l

v) ∆ C ≠ 0

4 4 M BC = M CB =

6 EI 6 EI 12 EI 4 4 4 4 ∆ C , M CD = M DC = − 2 ∆ C , VCD = VCB = 2 ∆C 2 l l l


13


Construct Stiffness Equation ql 2 EI EI + 8 θ B + 2 θCL + l l 12 i EI EI 2 θ B + 4 θCL + ∑i M 2i = 0 → l l

∑M

i 1

=0→−

∑M

i 3

=0→

EI ∆C = 0 l2 EI 0 + 6 2 ∆C = 0 l EI EI 4 θCR − 6 2 ∆ C = 0 l l EI EI EI EI 6 2 θ B + 6 2 θCL − 6 2 θCR + 24 3 ∆ C = 0 l l l l

i

∑V

i 4

=0→

i

z

0

+6

Elimination of θCL and θCR

- 2nd and 3rd equation 2

EI L EI EI EI EI θC = −( θ B + 3 2 ∆ C ) , 2 θCR = 3 2 ∆ C l l l l l

- 1st equation ql 2 EI EI EI − + 8 θ B + 2 θ CL + 6 2 ∆ C = 12 l l l 2 ql EI EI EI EI ql 2 EI EI − + 8 θB − ( θB + 3 2 ∆C ) + 6 2 ∆C = 0 → − +7 θB + 3 2 ∆C = 0 12 12 l l l l l l - 4th equation

EI EI EI EI θ B + 6 2 θ CL − 6 2 θ CR + 24 3 ∆ C = 2 l l l l EI EI EI EI EI EI EI 6 2 θ B − 3( 2 θ B + 3 3 ∆ C ) − 3(3 3 ∆ C ) + 24 3 ∆ C = 0 → 3 2 θ B + 6 3 ∆ C = 0 l l l l l l l

6

2.2.5 Analysis of a Beam with an Internal Hinge (2 DOFs System) q

C

B

A EI l z

DOF : θ B , ∆C

z

Analysis

D EI

EI

l

l

i) All fixed f M AB =

ql 2 ql 2 f , M BA =− 12 12


14

School of Civil, Urban & Geosystem Eng., SNU ii) θ B ≠ 0

M 1AB =

2 EI 4 EI 3EI 3EI 1 1 1 1 θ B , M BA = θ B , M BC = θ B , VBC = −VCB = − 2 θB l l l l

iii) ∆ C ≠ 0

2 2 M BC = − M DC =

z

3EI 3EI 3EI 2 2 2 2 ∆ C , VBC = −VCB = − 2 ∆ C , VCD = −VDC = 2 ∆C 2 l l l


∑M

=0→−

i 1

i

∑V

i 2

=0→

i

ql 2 EI EI + 7 θB + 3 2 ∆C = 0 12 l l EI EI 3 2 θB + 6 3 ∆C = 0 l l

θB =

ql 3 ql 4 , ∆C = − 66 EI 132 EI

2 EI R 3EI ∆ C 3 ∆C 3 ql 3 θC = − (− ) → θCR = =− l l l 2 l 264 EI 3 3 1 3∆ ql 3 ql ql 3 θCL = − θ B − =− + = 2 2L 132 EI 2 132 EI 264 EI

2.2.6 Beam with a Spring Support q

EI l z

C

B

A

D EI l

k

EI l

Analysis

i) All fixed f = M AB

ql 2 ql 2 f , M BA =− 12 12


15

School of Civil, Urban & Geosystem Eng., SNU ii) θ B ≠ 0

M 1AB =

2 EI 4 EI 3EI 3EI 1 1 1 1 θ B , M BA = θ B , M BC = θ B , − VBC = VCB = 2 θB l l l l

iii) ∆ C ≠ 0

k∆C 3EI ∆C l2 3EI 3EI 2 2 = − 2 ∆ C , VCD = −VDC = 2 ∆ C , VS2 = k∆ C l l

2 2 M BC = − M DC = 2 2 VBC = −VCB

z


∑ M 1i = 0 → − i

∑V

i 2

=0→

i

θB = z

ql 2 EI EI + 7 θB + 3 2 ∆C = 0 12 l l EI EI 3 2 θ B + (6 3 + k ) ∆ C = 0 l l

1+ α ql 3 1 ql 4 , ∆C = − (1 + 14α / 11) 132 EI 1 + 14α / 11 66 EI

6 EI l3

α→0

θB = z

where k = α

ql 3 ql 4 , ∆C = − 66 EI 132 EI

α→∞

θB =

ql 3 , ∆C = 0 84 EI


16

School of Civil, Urban & Geosystem Eng., SNU 2.2.7 Support Settlement

l

l

B

A EI

C

EI δ

z

DOF : θ B

z

Analysis

i) All fixed f M BA =

6 EI δ 3EI δ f , M BC =− l l l l

ii) θ B ≠ 0 1 M BA =

z

4 EI 3EI 1 θ B , M BC = θB l l

Construct the Equilibrium Equation

∑M i

i 1

=0→6

3δ EI δ EI δ EI EI −3 + 4 θB + 3 θB = 0 → θ B = − l l l l l l 7l

2.2.8 Temperature Change A

B T1 T2

θA = z

α(T2 − T1 ) α(T2 − T1 ) l , θB = − l 2h 2h

Fixed End Moment

4 EI θA + L 2 EI MB = θA + L MA =

2 EI α(T2 − T1 ) EI θB = L h 4 EI α(T2 − T1 ) EI θB = − L h


17

School of Civil, Urban & Geosystem Eng., SNU 2.3 Analysis of Frames 2.3.1 A Portal Frame without Sidesway

P

l/2 B

C EI2

z

DOF : θ B , θC

z

Analysis

EI1

EI1

A

D

i) All fixed Pl Pl 0 0 = , M CB =− M BC 8 8 ii) θ B ≠ 0 M 1AB =

2 EI1 4 EI1 1 θ B , M BA = θB l l

1 M BC =

4 EI 2 2 EI 2 1 θ B , M CB = θB l l

iii) θC ≠ 0

z

2 M BC =

2 EI 2 4 EI 2 2 θC , M CB = θC l l

2 M CD =

4 EI1 2 EI1 2 θC , M DC = θC l l


∑M

i B

=0→

4 EI1 4 EI 2 2 EI 2 Pl +( + )θ B + θC = 0 8 l l l

4 EI1 4 EI 2 Pl 2 EI 2 + θB + ( + ) θC = 0 8 l l l 1 Pl 2 − θ B = θC = 4 EI1 + 2 EI 2 8

∑M

i C

=0→−


18


Member End Forces

M AB =

2 EI1 2 EI1 Pl θB = − 4 EI1 + 2 EI 2 8 l

M BA =

4 EI1 4 EI1 Pl θB = − 4 EI1 + 2 EI 2 8 l

M BC =

2 EI 2 4 EI1 Pl 4 EI 2 Pl + θB + θC = 8 4 EI1 + 2 EI 2 8 l l

M CB = −

z

4 EI 2 4 EI1 Pl 2 EI 2 Pl + θB + θC = − 8 4 EI1 + 2 EI 2 8 l l

M CD =

4 EI1 4 EI1 Pl θC = 4 EI1 + 2 EI 2 8 l

M DC =

2 EI1 2 EI1 Pl θC = 4 EI1 + 2 EI 2 8 l

In case EI1 = EI 2

M AB = −

Pl Pl Pl Pl Pl , M BA = − , M BC = , M CB = − , M CD = 24 12 12 12 12

, M DC =

Pl 24

2.3.2 A Portal Frame without Sidesway – hinged suppoorts

P

l/2 B

C EI2

z

DOF : θ B , θC

z

Analysis

EI1

EI1

A

D

i) All fixed 0 = M BC

Pl Pl 0 , M CB =− 8 8



19

ii) θ B ≠ 0 3EI1 θB l 4 EI 2 2 EI 2 1 = θ B , M CB = θB l l

1 M BA = 1 M BC

iii) θC ≠ 0 2 EI 2 4 EI 2 2 θC , M CB = θC l l 3EI1 = θC l

2 M BC = 2 M CD

z


∑M

i B

=0→

Pl 3EI1 4 EI 2 2 EI 2 +( + )θ B + θC = 0 l l l 8

∑M

i C

=0→−

3EI1 4 EI 2 Pl 2 EI 2 + θB + ( + ) θC = 0 8 l l l

− θ B = θC = z

1 Pl 2 3EI1 + 2 EI 2 8

Member End Forces M AB = 0 3EI1 3EI1 Pl θB = − M BA = 3EI1 + 2 EI 2 8 l 2 EI 2 3EI1 Pl 4 EI 2 Pl + θB + θC = M BC = 8 3EI1 + 2 EI 2 8 l l 4 EI 2 3EI1 Pl 2 EI 2 Pl θB + θC = − M CB = − + 8 3EI1 + 2 EI 2 8 l l 3EI1 3EI1 Pl θC = M CD = 3EI1 + 2 EI 2 8 l M DC = 0

z

In case of EI1 = EI 2 M AB = 0 , M BA = −

3 3 3 3 Pl , M BC = Pl , M CB = − Pl , M CD = Pl , M DC = 0 40 40 40 40


20

School of Civil, Urban & Geosystem Eng., SNU 2.3.3 A Frame with an horizontal force

B

C

P

EI

A z

DOF : θ B , ∆

z

Analysis

i) All fixed : None fixed end moment ii) θ B ≠ 0 2 EI 4 EI 1 θ B , M BA = θB l l 3EI 6 EI 1 = θ B , VBA = 2 θB l l

M 1AB = 1 M BC

iii) ∆ ≠ 0 6 EI 6 EI 2 ∆ , M BA = 2 ∆ 2 l l 12 EI = 3 ∆ l

2 = M AB 2 VBA

z

Construct the stiffness equation

∑M ∑V

i

i B

=0→ (

4 EI 3EI 6 EI + )θ B + 2 ∆ = 0 l l l 6 EI 12 EI θB + 3 ∆ = P 2 l l

=P→

Pl 2 7 Pl 3 ,∆ = θB = − 8EI 48 EI z

Member end forces

M AB =

2 EI 6 EI 5 4 EI 6 EI 3 3EI 3 θ B + 2 ∆ = Pl , M BA = θ B + 2 ∆ = Pl , M BC = θ B = − Pl 8 8 8 l l l l l


21


2.3.4 A Portal Frame with an Unsymmetrical Load

a

P

B

A z

DOF : θ B , θC , ∆

z

Analysis

C

D


Pab 2 Pa 2b 0 , = − M CB l2 l2

ii) θ B ≠ 0 2 EI 4 EI 1 θ B , M BA = θB l l 4 EI 2 EI 6 EI 1 1 = θ B , M CB = θ B , VBA = 2 θB l l l

M 1AB = 1 M BC

iii) θC ≠ 0 2 EI 4 EI 2 θC , M CB = θC l l 4 EI 2 EI 6 EI 2 2 = θC , M DC = θC , VCD = 2 θC l l l

2 M BC = 2 M CD


22

School of Civil, Urban & Geosystem Eng., SNU iv) ∆ ≠ 0 6 EI 6 EI 3 3 ∆ , M CD = M DC = 2 ∆ , 2 l l 12 EI = 3 ∆ l

3 3 = M BA = M AB 3 3 = VCD VBA

z


∑ M Bi = 0 →

Pab 2 8EI 2 EI 6 EI + θB + θc + 2 ∆ = 0 2 l l l l

∑ M Ci = 0 → −

Pa 2b 2 EI 8 EI 6 EI + θB + θc + 2 ∆ = 0 2 l l l l

∑V

i

6 EI 6 EI 24 EI θ B + 2 θc + 3 ∆ = 0 2 l l l

=0→

l ∆ = − ( θ B + θC ) 4 Pa 2b 13EI EI + θB + θc = 0 2 l 2l 2l Pab 2 EI 13EI − 2 + θB + θc = 0 2l 2l l 1 Pab (a + 13b) θB = − l 84 EI 1 Pab (13a + b) θC = l 84 EI 1 Pab ∆= (b − a) 28 EI

a=

l 4

2.3.5 A Portal Frame with a Bracing (Vertical Load)

a B

P C

EI = 0, EA ≠ 0

z

DOF : θ B , θC , ∆

z

Analysis

A

D


Pab 2 Pa 2b 0 , = − M CB l2 l2


23

School of Civil, Urban & Geosystem Eng., SNU ii) θ B ≠ 0 2 EI 4 EI 1 θ B , M BA = θB l l 4 EI 2 EI 1 1 M BC = θ B , M CB = θB l l 6 EI 1 VBA = 2 θB l M 1AB =

iii) θC ≠ 0 2 EI 4 EI 2 θC , M CB = θC l l 4 EI 2 EI 2 2 M CD = θC , M DC = θC l l 6 EI 2 VCD = 2 θC l iv) ∆ ≠ 0 2 M BC =

6 EI M = M = 2 ∆, l 6 EI 3 3 M CD = M DC = 2 ∆, l 12 EI 3 3 VBA = VCD = 3 ∆ l 3 AB

3 BA

ABD = z

∆ ∆ 2

EA ∆ EA ∆ 1 EA ∆ (C) → VBD = = VBD = 2l 2 2 2l 2 2l 2


Pab 2 8EI 2 EI 6 EI + θB + θc + 2 ∆ = 0 2 l l l l 2 Pa b 2 EI 8 EI 6 EI ∑ M Ci = 0 → − l 2 + l θB + l θc + l 2 ∆ = 0 6 EI 6 EI 24 EI θ B + 2 θc + 3 (1 + α)∆ = 0 ∑V i = 0 → 2 l l l EAl 2 1 l ∆=− (θ B + θC ) , α = 1+ α 4 48 2 EI

∑ M Bi = 0 →

z

Solution for b = 3a θB = −

40 + 52α Pl 2 16 + 28α Pl 2 3 Pl 3 , θC = − , ∆= 256(7 + 10α ) EI 256(7 + 10α) EI 128(7 + 10α) EI

For a w × h rectangular section and l = 20h , α = 50 2 . Pl 2 Pl 2 Pl 3 θ B = −0.0203 , θC = 0.0109 ∆ = 0.3282 × 10 −4 EI EI EI


24


Performance

with Bracing ( α = 50 2 )

Response

w/o bracing (α = 0)

Ratio(%)

θB (× Pl 2 / EI )

-0.0203

-0.0223

91.03

θC (× Pl 2 / EI )

0.0109

0.0089

122.47

∆ (× Pl 3 / EI )

0.3282×10-4

0.0033

0.99

ΜΑΒ (Pl)

-0.0404

-0.0248

162.90

ΜΒΑ (Pl)

-0.0810

-0.0694

116.71

ΜCD (Pl)

0.0438

0.0554

79.06

ΜDC (Pl)

0.0220

0.0376

58.51

ΜP (Pl)

0.1158

0.1216

95.23

ABD (P)

0.0788

Pmax (Pall)*

0.0720

0.0685

105.1

Pmax/vol.

0.0163

0.0228

71.5

-

-

*) Pall = σ all wh , M all = Pall h / 6 Unbalanced shear force in the columns =

6 EI (θ B + θC ) = 0.0564 P l2

The bracing carries 99 % of the unbalanced shear force between the two columns.

2.3.6 A Portal Frame with a Bracing (Horizontal Load)

B

P

C

EI = 0, EA ≠ 0

A z

DOF : θ B , θC , ∆

z

Analysis

D

i) All fixed: No fixed end forces ii)-iv) the same as the previous case


25



8EI 2 EI 6 EI θB + θc + 2 ∆ = 0 l l l EI EI EI 2 8 6 ∑ M Ci = 0 → l θB + l θc + l 2 ∆ = 0 6 EI 6 EI 24 EI ∑V i = 0 → l 2 θB + l 2 θc + l 3 (1 + α)∆ = P

∑M

i B

=0→

5 5 θ B = θC , ∆ = − θ B l = − θC l 3 3 z

Solution 1 Pl 2 5 Pl 3 θ B = θC = − , ∆= (28 + 40α) EI 3(28 + 40α) EI

For α = 50 2 , θ B = θC = −0.3501 × 10− 3 z

Pl 2 Pl 3 , ∆ = 0.5835 × 10− 3 EI EI

Performance

Response

with Bracing ( α = 50 2 )

w/o bracing (α = 0)

Ratio(%)

θB (× Pl 2 / EI )

− 0.3501 × 10−3

− 0.3571 × 10−1

0.98

θC (× Pl 2 / EI )

− 0.3501 × 10−3

− 0.3571 × 10−1

0.98

∆ (× Pl 3 / EI )

0.5835 × 10−3

0.5952 × 10−1

0.98

ΜΑΒ (Pl)

0.2801 × 10−2

0.2857

0.98

ΜΒΑ (Pl)

0.2101 × 10−2

0.2143

0.98

ΜCD (Pl)

0.2101 × 10−2

0.2143

0.98

ΜDC (Pl)

0.2801 × 10−2

0.2857

0.98

ABD (P)

1.4004

-

-

Pmax(Pall)*

0.7141

0.0292

2448

Pmax/vol.

0.1617

0.0097

1670

*) Governed by ABD for the structure with bracing, and by MDC for the structure without bracing. Pall = σ all wh , M all = Pall h / 6 The bracing carries about 99% of the external horizontal load.


26

School of Civil, Urban & Geosystem Eng., SNU 2.3.7 A Portal Frame with a Spring

a

P

B

C k

A z z

D

DOF : θ B , θC , ∆ Analysis

k∆

iv) ∆ ≠ 0 6 EI 6 EI 3 3 ∆ , M CD = M DC = 2 ∆ , 2 l l 24 EI = 3 ∆ , VS3 = k∆ l

3 3 M AB = M BA = 3 3 VBA = VCD

z


Pab 2 8EI 2 EI 6 EI ∑ M = 0 → l 2 + l θ B + l θc + l 2 ∆ = 0 Pa 2b 2 EI 8 EI 6 EI i M = 0 → − + θB + θc + 2 ∆ = 0 ∑ C 2 l l l l 6 EI 6 EI 24 EI θ B + 2 θc + 3 ∆ = − k∆ ∑V i = 0 → 2 l l l i B

6 EI 6 EI 24 EI θ B + 2 θc + ( 3 + k ) ∆ = 0 2 l l l z

Deformed Shapes (

∆S = 0.41 ) ∆

without a spring

with a spring ( k = 24

EI ) l3


27

School of Civil, Urban & Geosystem Eng., SNU 2.3.8 A Portal Frame Subject to Support Settlement

B

C

A

D δ

z

DOF : θ B , θC , ∆

z

Analysis

i) All fixed 6 EI δ l2 ii)-iv) the same as the previous problem 0 0 M BC = M CB =

z

Construct the Stiffness Equation 6 EI 8EI 2 EI 6 EI ∑ M Bi = 0 → l 2 δ + l θB + l θc + l 2 ∆ = 0 6 EI 2 EI 8EI 6 EI ∑ M Ci = 0 → l 2 δ + l θB + l θc + l 2 ∆ = 0 6 EI 6 EI 24 EI θ B + 2 θc + 3 ∆ = 0 ∑V i = 0 → 2 l l l


28

School of Civil, Urban & Geosystem Eng., SNU 2.3.9 A Portal Frame with Unsymmetrical Supports

l/2 B

A z

DOF : θ B , θC , ∆

z

Analysis

P C

D

i) All fixed 0 M BC =

Pl Pl 0 , M CB =− 8 8

ii) θ B ≠ 0 3EI θB l 4 EI 2 EI 1 1 M BC = θ B , M CB = θB l l 3EI 1 VBA = 2 θB l 1 M BA =

iii) θC ≠ 0 2 EI 4 EI 2 θC , M CB = θC l l 4 EI 2 EI 2 2 M CD = θC , M DC = θC l l 6 EI 2 VCD = 2 θC l 2 M BC =

iv) ∆ ≠ 0 3 M BA =

3EI ∆, l2

6 EI ∆ , l2 3EI 12 EI 3 = 3 ∆ , VCD = 3 ∆ l l

3 3 M CD = M DC = 3 V BA


29



Pl 7 EI 2 EI 3EI + θB + θc + 2 ∆ = 0 l l l 8 Pl EI EI EI 2 8 6 ∑ M Ci = 0 → − 8 + l θB + l θc + l 2 ∆ = 0 3EI 6 EI 15 EI θ B + 2 θc + 3 ∆ = 0 ∑V i = 0 → 2 l l l l ∆ = − (θ B + 2θC ) 5 2 1 Pl 1 9 Pl 2 1 Pl 3 θB = − , θC = , ∆=− 44 EI 44 8 EI 176 EI

∑M

z

i B

=0→

Load Location that Causes No Sidesway

P

a

C

B

A

D

l ∆ = − (θ B + 2θC ) = 0 → θ B = −2θC 5 - Stiffness equation

∑ M Bi = 0 →

Pab 2 7 EI 2 EI + θB + θc = 0 , 2 l l l Pab 2 12 EI − θC = 0 l2 l

,

∑ M Ci = 0 → − −

Pa 2b 2 EI 8EI + θB + θc = 0 2 l l l

Pa 2b 4 EI + θc = 0 l2 l

Pab 2 Pa 2b 3 = → b = 3a l2 l2


30

School of Civil, Urban & Geosystem Eng., SNU 2.3.10 A Frame with a Skewed Member

P

B

C

A

z

DOF : θ B , ∆

z

Analysis

0 i) All fixed : M BC =

2 11 Pl Pl 3 0 + = Pl , VBC =− P 2 16 8 16 16

ii) θ B ≠ 0 M 1AB =

2 EI EI EI 3EI EI 1 1 1 θB = 2 θ B , M BA =2 2 θ B , M BC = θ B , VBA = 3 2 θB l l l l 2l

1 VBC =−

3 2 EI θB 2 l2

3 (2 2

EI 1 2 2 EI θB =3 θB l l 2 2 l2

EI EI EI 1 θB + 2 θB ) = 3 2 θB l l l 2l

iii) ∆ ≠ 0 2 2 M BA = M AB = 2 VBC =

3EI ∆ 3 2 EI EI 6 EI ∆ 3EI 2 2 = 2 ∆ M BC =− =− ∆ , VBA =3 2 3 ∆ , 2 l l l 2 l 2l 2l 2l

3 EI ∆ 2 l3

3 (3

2 EI 1 1 3 EI ∆ = ∆ 2 l 2 2 l3 2 l

EI EI EI 1 ∆ + 3 2 ∆) = 3 2 3 θB 2 l l l 2l



31

Construct the stiffness equation

3 2 EI EI 3 ) 2 ∆ = − Pl θ B + (3 − 2 l l 16 3 3 EI EI 2 11 ∑V i = 0 → (3 − 2 2) l 2 θB + (3 2 + 2 ) l 3 ∆ = 2 16 P

∑M

i B

= 0 →(2 2 + 3)

EI EI θ B + 0.8787 2 ∆ = −0.1875 Pl l l EI EI 0.8787 2 θ B + 5.7426 3 ∆ = 0.4861P l l

5.8284

Pl 2 Pl 3 θ B = −0.0460 , ∆ = 0.0917 EI EI z

Results

- Deformed shape

- Moment diagram

- Shear force diagram



32

Chapter 3 Iterative Solution Method & Moment Distribution Method


33

School of Civil, Urban & Geosystem Eng., SNU 3.1 Solution Method for Linear Algebraic Equations 3.1.1 Direct Method – Gauss Elimination a11 X 1 + a12 X 2 + L + a1i X i a 21 X 1 + a 22 X 2 + L + a 2i X M a i1 X 1 + ai 2 X 2 + L + aii X i M a n1 X 1 + a n 2 X 2 + L + a ni X i

+ L + a1n X n = b1  + L + a 2 n X n = b2   n  →  ∑ aij X j = bi for i = 1L n + L + a in X n = bi  j =1   + L + a nn X n = bn 

or in a matrix form

[ A]( X) = (b) By multiplying

ai1 to the first equation and subtracting the resulting equation from the i-th a11

equation for 2 ≤ i ≤ n , the first unknown X1 is eliminated from the second equation as follows. a11 X 1 + a12 X 2 + L + a1i X i ( 2) a 22 X 2 + L + a 2( 2i ) X M ( 2) ( 2) ai 2 X 2 + L + aii X i M ( 2) ( 2) an2 X 2 + L + an2 X i

where aij( 2) = aij −

ai1 a1 j a11

+ L + a1n X n = b1 + L + a 2( 2n) X n = b2( 2 ) + L + ain( 2) X n = bi( 2 ) ( 2) + L + a nn X n = bn( 2 )

. Again, the second unknown X2 is eliminated from the third equa-

ai(22) tion by multiplying ( 2) to the second equation and subtracting the resulting equation from a 22

the i-th equation for 3 ≤ i ≤ n .

The aforementioned procedures are repeated until the last

unknown remains in the last equation. a11 X 1 + a12 X 2 + L + a1i X i + L + a1n X n ( 2) a 22 X 2 + L + a 2( 2i ) X + L + a 2( 2n) X n M M (i ) (i ) aii X i + L + ain X n M (n) a nn X n

= b1 = b2( 2 ) M = bi( i ) M (n) = bn


34


a

=a

( k −1) ij

( n −1) n −1, n −1

−

ai(,kk−−11) a k( k−−1,1j)

k ≤ i, j ≤ n , and a ij1 = aij . Once the system matrix is tria kk−−11,k −1 angularized, the solution of the given system is easily obtained by the back-substitution. bn( n ) (n) ( n) X n = (n) a nn X n = bn → a nn

where a

(k ) ij

X n −1 + a

( n −1) n −1, n

Xn = b

( n −1) n −1

→

X n −1 =

bn( n−1−1) − a n( n−−1,1n) X n a nn−−12,n −1 i +1

aii( i ) X i + L + a i(,in) X n = bi(i ) →

Xi =

bi( i ) − ∑ aik(i ) X k k =n i −1 ii

a

for 1 ≤ i ≤ n − 1

3.1.2 Iterative Method – Gauss-Jordan Method

A system of linear algebraic equations may be solved by iterative method. For this purpose, the given system is rearranged as follows. X1 =

b1 − (a12 X 2 + L + a1n X n ) a11

X2 =

b2 − (a 21 X 1 + a 23 X 3 + L + a 2 n X n ) a 22

Xi = Xn =

bi − (ai1 X 1 + L + ai ,i −1 X i −1 + ai ,i +1 X i −1 + L + ain X n ) aii bn − (a n1 X 1 + L + a n ,n −1 X n −1 ) a nn

Suppose we substitute an approximate solution ( X) k −1 into the right-hand side of the above equation, a new approximate solution ( X) k , which is not the same as ( X) k −1 , is obtained. This procedure is repeated until the solution converges. ( X i )k =

n 1 (bi − ∑ aij ( X j ) k −1 ) aii j =1 i≠ j

where the subscript k denotes the iterational count.


35

School of Civil, Urban & Geosystem Eng., SNU 3.1.3 Iterative Method – Gauss-Siedal Method

When we calculate a new Xi value in the k-th iteration of Gauss-Jordan iteration, the values of X 1 ,L, X i −1 are already updated, and we can utilize the updated values to accelerate conver-

gence rate, which leads to the Gauss-Siedal Method. ( X1)k =

b1 − (a12 ( X 2 ) k −1 + L + a1n ( X n ) k −1 ) a11

( X 2 )k =

b2 − (a 21 ( X 1 ) k + a 23 X 3 ) k −1 + L + (a 2 n X n ) k −1 ) a 22

( X i )k = ( X n )k =

bi − (ai1 ( X 1 ) k + L + ai ,i −1 ( X i −1 ) k + ai ,i +1 ( X i +1 ) k −1 + L + ain ( X n ) k −1 ) aii bn − (a n1 ( X 1 ) k + L + a n ,n −1 ( X n −1 ) k ) a nn ( X i )k =

i −1 1 (bi − ∑ aij ( X j ) k − aii j =1 i >1

n

∑a

j =i +1 i
ij

( X j ) k −1 )

3.1.4 Example

30

35.6 B

A EI

4klf C

1.5EI

D EI

3@30=90 z

Stiffness Equation

3EI 6 EI 3EI + )θ B + θC = 0 l l l 3EI 6 EI 3EI − 133.3 + 450.0 + θB + ( + )θC = 0 l l l − 168.7 + 133.3 + (

For the simplicity of derivation,

EI EI θB → θB , θ C → θ C . The stiffness equation becomes l l − 35.4 + 9θ B + 3θ C = 0 316.7 + 3θ B + 9θ C = 0


36


Gauss-Jordan Iteration

1 (35.4 − 3(θ C ) k −1 ) 9 1 (θ C ) k = (316.7 − 3(θ B ) k −1 ) 9

(θ B ) k =

z

Gauss-Siedal Iteration

1 (35.4 − 3(θ C ) k −1 ) 9 1 (θ C ) k = (316.7 − 3(θ B ) k ) 9

(θ B ) k =

Gauss-Seidal

Gauss-Jordan

GAUSS-SIEDAL ITERATION ====================== ***** Iteration 1***** X(1) = 0.3933334E+01 X(2) = -0.3650000E+02 ERROR = 0.1000000E+01 ***** Iteration 2***** X(1) = 0.1610000E+02 X(2) = -0.4055556E+02 ERROR = 0.2939146E+00 ***** Iteration 3***** X(1) = 0.1745185E+02 X(2) = -0.4100617E+02 ERROR = 0.3197497E-01 ***** Iteration 4***** X(1) = 0.1760206E+02 X(2) = -0.4105624E+02 ERROR = 0.3544420E-02 ***** Iteration 5***** X(1) = 0.1761875E+02 X(2) = -0.4106181E+02 ERROR = 0.3937214E-03

GAUSS-Jordan ITERATION ====================== ***** Iteration 1***** X(1) = 0.3933334E+01 X(2) = -0.3518889E+02 ERROR = 0.1000000E+01 ***** Iteration 2***** X(1) = 0.1566296E+02 X(2) = -0.3650000E+02 ERROR = 0.2971564E+00 ***** Iteration 3***** X(1) = 0.1610000E+02 X(2) = -0.4040988E+02 ERROR = 0.9044394E-01 ***** Iteration 4***** X(1) = 0.1740329E+02 X(2) = -0.4055556E+02 ERROR = 0.2971564E-01 ***** Iteration 5***** X(1) = 0.1745185E+02 X(2) = -0.4098999E+02 ERROR = 0.9812154E-02

****** MOMENT ******

****** MOMENT ******

MBA MBC MCB MCD

=-115.84375 = 115.82707 =-326.81460 = 326.81458

MBA MBC MCB MCD

=-116.34444 = 115.04115 =-326.88437 = 327.03004


37

School of Civil, Urban & Geosystem Eng., SNU 3.2 Moment Distribution Method 30

35.6 B

A EI

4klf C

1.5EI

D EI

3@30=90

z

At the Joint B

- Moment distribution MB = (

1 = M BA

MB 3EI AB 4 EI BC 1 1 + + M BC → θB = )θ B = M BA 3EI AB 4 EI BC L AB LBC + L AB LBC 3EI AB L AB

3EI AB L AB θB = M = DBA M B 3EI AB 4 EI BC B + L AB LBC

4 EI BC 4 EI BC LBC 1 = θB = M BC M = DBC M B 3EI AB 4 EI BC B LBC + L AB LBC 2 EI BC 1 1 - Moment carry over to joint C: M CB = θ B = DBC M B 2 LBC z

At the Joint C

- Moment distribution MC = (

2 M CB =

2 = M CD

4 EI BC 3EI CD MC 2 2 + )θ C = M CB + M CD → θC = 4 EI BC 3EI CD LBC LCD + LBC LCD 4 EI BC LBC

4 EI BC LBC θC = M = DCB M C 4 EI BC 3EI CD C + LBC LCD

3EI CD LCD

3EI CD LCD θC = M = DCD M C 4 EI BC 3EI CD C + LBC LCD

2 - Moment carry over to joint B: M BC =

2 EI BC 1 θ C = DCB M C LBC 2


38


Stiffness Equation in terms of Moment at Joints

1  1 DCB M C = 0 M B = 35.4 − DCB M C  2 2 → 1 1 316.7 + DBC M B + M C = 0  M C = - 316.7 − DBC M B 2  2 − 35.4 + M B +

- Gauss-Siedal Approach 1 DCB ( M C ) k −1 2 1 ( M C ) k = − 316.7 − DBC ( M B ) k 2

( M B ) k = 35.4 −

- Gauss-Jordan Approach 1 DCB ( M C ) k −1 2 1 ( M C ) k = − 316.7 − DBC ( M B ) k −1 2

( M B ) k = 35.4 −

z

For the given structure

D BA = DCD = z

1 , 3

D BC = DCB =

2 3

Incremental form for the Gauss-Siedal Method

- For k = 1 ( M B ) 0 = ( M C ) 0 = 0 because we assume all degrees of freedom are fixed for step 0. 1 DCB ( M c ) 0 = 35.4 → (∆M B )1 = 35.4 2 1 12 ( M C )1 = −316.7 − DBC ( M B )1 = −316.7 − 35.4 → (∆M C )1 = −328.5 2 23 ( M B )1 = 35.4 −

- For k > 1 1 1 1 DCB ( M c ) k −1 = 35.4 − DCB ( M c ) k − 2 − DCB (∆M c ) k −1 2 2 2 1 1 = ( M B ) k −1 − DCB (∆M c ) k −1 → (∆M B ) k = − DCB (∆M c ) k −1 2 2 1 1 1 ( M C ) k = −316.7 − DBC ( M B ) k = −316.7 − DBC ( M B ) k −1 − DBC (∆M B ) k 2 2 2 1 1 = ( M C ) k −1 − DBC (∆M B ) k → (∆M C ) k = − DBC (∆M B ) k 2 2 ( M B ) k = 35.4 −



39

- Iteration 1 (M B ) 0 = 0 , (M C ) 0 = 0 ( M B ) 1 = 35.4 −

1 DCB ( M C ) 0 = 35.4 → (∆M B ) 1 = 35.4 2

f + DBA (∆M B )1 = −168.7 + 0.33 × 35.4 = −168.7 + 11.8 → (∆M BA )1 = 11.8 ( M BA )1 = M BA f ( M BC )1 = M BC + DBC (∆M B )1 = 133.3 + 0.67 × 35.4 = 133.3 + 23.6 → (∆M BC )1 = 23.6

( M C )1 = −316.7 − f ( M CB )1 = M CB +

1 12 DBC (∆M B )1 = −316.7 − 35.4 = −328.5 → (∆M c )1 = −328.5 2 23 1 DBC (∆M B )1 + DCB (∆M C )1 = −133.3 + 11.8 − 219.0 → 2 (∆M CB )1 = 11.8 − 219.0

f ( M CD )1 = M CD + DCD (∆M C )1 = 450 − 0.33 × 328.5 = 450 − 109.5 → (∆M CD )1 = −109.5

- Iteration 2

(∆M BA ) 2 = DBA (∆M B ) 2 = 36.5  1 1 (∆M B ) 2 = − DCB (∆M c )1 = 109.5 → (∆M BC ) 2 = DCB (∆M c )1 + DBC (∆M B ) 2 2 2   = −109.5 + 73.0 (∆M ) = 1 D (∆M ) + D (∆M ) CB 2 BC B 2 CB C 2  1 2 (∆M C ) 2 = − DBC (∆M B ) 2 = −36.5 →  = 36.5 − 24.3 2  (∆M CD ) 2 = DCD (∆M C ) 2 = −12.2

- Iteration 3

(∆M BA ) 3 = D BA ( ∆M B ) 3 = 4.1  1 1 (∆M B ) 3 = − DCB (∆M c ) 2 = 12.2 → (∆M BC ) 3 = DCB (∆M c ) 2 + D BC (∆M B ) 3 2 2   = −12.2 + 8.2 ( ∆M ) = 1 D (∆M ) + D (∆M ) = 4.1 − 2.7 1 CB 3 BC B 3 CB C 3 (∆M C ) 3 = − D BC (∆M B ) 3 = −4.1 →  2 2 ( ∆M CD ) 3 = DCD (∆M C ) 3 = −1.4

- Final Moments f M BA = M BA + ∑ (∆M BA ) k = −168.7 + 11.8 + 36.5 + 4.1 = −116.3 k

M BC = M

f BC

M CB = M

f CB

M CD = M

f CD

+ ∑ (∆M BC ) k = 133.3 + 23.6 + (−109.5 + 73.0) + (−12.2 + 8.2) = 116.4 k

+ ∑ (∆M CB ) k = −133.3 + (11.8 − 219.0) + (36.5 − 24.3) + (4.1 − 2.7) = −326.9 k

+ ∑ (∆M CD ) k = 450.0 − 109.5 − 12.2 − 1.4 = 326.9 k


40

School of Civil, Urban & Geosystem Eng., SNU 0.33

0.66

0.67

-168.7

133.3

-133.3

11.8

23.6

11.8

-109.5

-219.0

73.0

36.5

-12.2

-24.3

8.2

4.1

-1.4

-2.7

-1.4

0.5

0.9

-326.9

326.9

-115.8

115.9

36.5

4.1

z

0.33 450.0

-109.5

-12.2

Incremental form for the Gauss-Jordan Method

- For k = 1 ( M B ) 0 = ( M C ) 0 = 0 because we assume all degrees of freedom are fixed for step 0. 1 DCB ( M c ) 0 = 35.4 → (∆M B )1 = 35.4 2 1 ( M C )1 = −316.7 − DBC ( M B ) 0 = −316.7 → (∆M C )1 = −316.7 2 ( M B )1 = 35.4 −

- For k > 1 1 1 1 DCB ( M c ) k −1 = 35.4 − DCB ( M c ) k − 2 − DCB (∆M c ) k −1 2 2 2 1 1 = ( M B ) k −1 − DCB (∆M c ) k −1 → (∆M B ) k = − DCB (∆M c ) k −1 2 2 1 1 1 ( M C ) k = −316.7 − DBC ( M B ) k −1 = −316.7 − DBC ( M B ) k − 2 − DBC (∆M B ) k −1 2 2 2 1 1 = ( M C ) k −1 − DBC (∆M B ) k −1 → (∆M C ) k = − DBC (∆M B ) k −1 2 2 ( M B ) k = 35.4 −

- Iteration 1 (M B ) 0 = 0 , (M C ) 0 = 0 ( M B ) 1 = 35.4 −

1 DCB ( M C ) 0 = 35.4 → (∆M B ) 1 = 35.4 2

f + D BA ( M B ) 1 = −168.7 + 0.33 × 35.4 = −168.7 + 11.8 → (∆M BA ) 1 = 11.8 ( M BA ) 1 = M BA f ( M BC ) 1 = M BC + D BC ( M B ) 1 = 133.3 + 0.67 × 35.4 = 133.3 + 23.6 → (∆M BC ) 1 = 23.6


41

School of Civil, Urban & Geosystem Eng., SNU ( M C )1 = −316.7 −

1 DBC ( M B ) 0 = −316.7 = −316.7 → (∆M c )1 = −316.7 2

f ( M CB )1 = M CB + DCB ( M C )1 = −133.3 − 0.67 × 316.7 = −133.3 − 212.2 →

(∆M CB )1 = −212.2 f ( M CD )1 = M CD + DCD ( M C )1 = 450 − 0.33 × 316.7 = 450 − 104.5 → (∆M CD )1 = −104.5

- Iteration 2

(∆M BA ) 2 = DBA (∆M B ) 2 = 35.0  1 1 (∆M B ) 2 = − DCB (∆M c )1 = 106.1 → (∆M BC ) 2 = DCB (∆M c )1 + DBC (∆M B ) 2 2 2   = −106.1 + 71.1 (∆M ) = 1 D (∆M ) + D (∆M ) CB 2 BC B 1 CB C 2  1 2 (∆M C ) 2 = − DBC (∆M B )1 = −11.8 →  = 11.8 − 7.9 2  (∆M CD ) 2 = DCD (∆M C ) 2 = −3.9

- Iteration 3

(∆M BA ) 3 = DBA (∆M B ) 3 = 1.3  1 1 (∆M B ) 3 = − DCB (∆M C ) 2 = 4.0 → (∆M BC ) 3 = DCB (∆M C ) 2 + DBC (∆M B ) 3 2 2   = − 4 .0 + 2 .7 (∆M ) = 1 D (∆M ) + D (∆M ) CB 3 BC B 3 CB C 3  1 2 (∆M C ) 3 = − DBC (∆M B ) 2 = −35.6 →  = 35.6 − 23.9 2  (∆M CD ) 3 = DCD (∆M C ) 3 = −11.7

0.33

0.66

0.67

-168.7 11.8

35.0 1.3 4.0 0.2 0.5 -115.9

133.3 23.6 -106.1 71.1 -4.0 2. 7 -12.0 8.0 -0.5 0.3 -1.4 0.9 115.9

0.33 -133.3 -212.2 11.8 -7. 9 35.6 -23.9 1.4 -0.9 4.0 -2.7 0.2 0.1 -328.0

450.0 -104.5

-3.9 -11.7 -0.5 -1.3 0.1 328.2


42

School of Civil, Urban & Geosystem Eng., SNU 3.3 Example - MDM for a 4-span Continuous Beam 100 A

DCD = 4 z

EI L

(4

1.5EI, L

1.5EI L

EI 1.5 EI 1.5 EI +4 ) = 0.4 , D BC = 4 L L L

(4

E

D

C

B EI, L

D BA = 4

50 1.5EI, L

(4

1.5EI 1.5 EI 1.5 EI +4 ) = 0.5 , D DC = 4 L L L

EI, L

EI 1.5 EI 1.5EI +4 ) = 0.6 , DCB = 4 L L L (4

(4

1.5EI 1.5EI +4 ) = 0.5 , L L

1.5EI EI EI + 3 ) = 0.67 , DDE = 3 L L L

(4

1.5 EI EI + 3 ) = 0.33 L L

Gauss-Siedal Approach 0.4 0.0 -25.0

-10.4

-1.1 36.5

0.0 -50.0

-20.8

-2.1 -72.9

0.6 125 -75 40.6

11.3 -31.1 3.9 1.3 -3.1 72.9

0.5 -125 -37.5 81.3

22.5 -15.6 7.8 2.6 -63.9

0.67

0.5 0.0

0.0

81.3 -45.0 22.5

40.6 -90.0 11.3

7.8 -5.1 2.6

3.9 -10.2 1.3 -0.9 -44.0

64.1

0.33 93.8

-44.3

-5.0 -0.4 44.1


43


Gauss-Jordan Approach 0.4 0.0

0.0 -50

-25.0 -12.5 -6.3 -6.9 -3.5 -2.0 -1.0 -1.1 -0.6 -36.4

-0.3 -72.8

0.6 125 -75 31.3 -18.8 17.3 -10.4 5.0 -3.0 2.8 -1.7 0.8 -0.5 72.8

0.5 -125 62.5 -37.5 34.5 -9.4 10.0 -5.2 5.5 -1.5 1.6 -0.9 1.0 -64.4

0.67

0.5 0.0 62.5 -31.4 34.5 -10.5 10.0 -5.8 5.5 -1.7 1.6 -1.0 1.0 64.7

0.0 -62.8 31.3 -21.0 17.3 -11.6 5.0 -3.4 2.8 -1.9 0.8 -0.5 -44.0

0.33 93.8 -31.0

-10.3 -5.7 -1.6 -0.9 -0.3 44.0


44

School of Civil, Urban & Geosystem Eng., SNU 3.4 Direct Solution Scheme by Partitioning 20 L/3

B

C

A

z

D

Slope deflection (Stiffness) Equation

 8EI  L  2 EI   L  6 EI  L

2 EI L 8EI L 6 EI L

6 EI  L  θ B   88.8    [K θθ ] (K θ∆ ) (Θ)   (P)  6 EI     θC  +  − 44.4  = 0 →   ∆  +  0  = 0 ( ) K K L     ∆ θ ∆∆      24 EI  ∆   0.0  L 

[K θθ ](Θ) = −(P) − (K θ∆ )∆ → (Θ) = −[K θθ ] −1 (P) − [K θθ ] −1 (K θ∆ )∆ = (Θ) P + (Θ) ∆ (K ∆θ )(Θ) P + (K ∆θ )(Θ) ∆ + K ∆∆ ∆ = 0 z

Direct Solution Procedure by Partitioning -

Assume ∆ = 0 and calculate (Θ)P. ∆ Assume an arbitrary ∆ = and calculate ( Θ ) ∆ . α ∆ ( Θ) By linearity, ( Θ ) ∆ = α Calculate α by the second equation by ∆ and ( Θ ) ∆ . (K ∆θ )(Θ) P α=− K ∆∆ ∆ + (K ∆θ )(Θ ) ∆

-

Obtain (Θ) by

(Θ ) = (Θ ) P + α ( Θ ) ∆ .


45

School of Civil, Urban & Geosystem Eng., SNU 3.5 Moment Distribution Method for Frames z

z

Solution Procedure -

Assume there is no sideway and do the MDM.

-

Perform the MDM again for an assumed sidesway.

-

Adjust the Moment obtained by the second MDM to satisfy the second equation.

-

Add the adjusted moment to the moment by the first MDM.

First MDM – with no Sidesway L/3

20

B 0.5

0.5 C

0.5

-44.4

-44.4

-22.2

16.6

33.3

8.3

-4.2

1.1

2.1

-0.6

-0.6

-0.3

-53.3

53.3

0.2

0.2

-35.5

35.5

-44.4

-8.3

0.5

88.8

33.3

2.1

A

D

MAB = -53.3/2 = -26.7, MDC = -35.5/2 = 17.8, VAB = 2.67, VDC = -1.78, VP = 0.89 z

Second MDM – with an Arbitrary Sidesway B

C 10.0

10.0 -5.0

-2.5

-1.9

-3.8

1.0

0.5

-0.2

-0.3

-0.3

0.1

0.1

-5.9

5.9

6.1

-6.0

-5.0

1.0

A

-3.8

D

MAB = 10 – 3.9/2.0 = 8.1, MDC = 10 – 4.1/2.0 = 7.9, VAB =-0.47, VDC = -0.46, V ∆ = -0.93


46


Shear Equilibrium Condition (the 2nd equation)

Sidesway Only (2nd MDM)

No Sidesway (1st MDM)

0.93

0.89 53.3

35.5 2.67

26.7

5.9

6.1 0.47

1.78

17.8

8.1

0.46

7.9

α = -0.89/(-0.93) = 0.97

Total Moment = 1st Moment + α 2nd Moment


47


Chapter 4 Energy Principles Principle of Minimum Potential Energy and Principle of Virtual Work

Mg

h

P


48


Read Chapter 11 (pp.420~ 428) of Elementary Structural Analysis 4th Edition by C .H. Norris et al very carefully.

In this note an overbarred variable denotes a virtual quantity.

The vir-

tual displacement field should satisfy the displacement boundary conditions of supports if specified. For beam problems, displacement boundary conditions include boundary conditions for rotational angle. Variables with superscript e denote the exact solution that satisfies the equilibrium equation(s).

4.1 Spring-Force Systems

∆ P

z

Total Potential energy The energy required to return a mechanical system to a reference status ∆ 0 1 Π int = − ∫ k (∆ − u )du = ∫ k (∆ − u )du = k∆2 , Π ext = − P∆ 2 ∆ 0 Π total = Π int + Π ext =

z

1 2 k∆ − P∆ 2

Equilibrium Equation k∆e=P

z

Principle of Minimum Potential Energy for an arbitrary displacement ∆ = ∆e + ∆ .

1 k (∆e + ∆ ) 2 − P(∆e + ∆ ) 2 1 1 = k (∆e ) 2 + k∆e ∆ + k ( ∆ ) 2 − P(∆e + ∆ ) 2 2 1 1 = k (∆e ) 2 − P∆e + k ( ∆ ) 2 + ∆ (k∆e − P) 2 2 1 1 = k (∆e ) 2 − P∆e + k ( ∆ ) 2 2 2 1 e e = Π total + k ( ∆ ) 2 ≥ Π total 2

Π total =

In the above equation, the equality sign holds if and only if ∆ = 0 .

Therefore the total

potential energy of the spring-force system becomes minimum when displacement of spring satisfies the equilibrium equation.


49

School of Civil, Urban & Geosystem Eng., SNU 4.2 Beam Problems q

w

any type of support z

w

Potential Energy of a Beam l

Π int =

l

l

l

l

1 d 2w 2 EI ( ) dx − ∫ qwdx 2 ∫0 dx 2 0

Equilibrium Equation EI

z

any type of support

d 2w 2 1 M2 1 = dx EI ( ) dx , Π ext = − ∫ qwdx 2 ∫0 EI 2 ∫0 dx 2 0

Π total = Π int + Π ext = z

we

d 2M e d 4 we = − q or EI =q dx 2 dx 4

Principle of Minimum Potential Energy for a virtual displacement w = w e + w . 1 d 2 (we + w ) d 2 (we + w ) ( )dx − ∫ ( w e + w )qdx EI 2 2 ∫ 20 dx dx 0 l

Πh =

l

l

=

l

1 d 2 we d 2 we ( )dx − ∫ w e qdx + EI 2 ∫0 dx 2 dx 2 0 l

l

l

1 d 2w d 2w d 2w d 2 we + ( )dx ∫ ( 2 EI )dx − ∫ w qdx EI 2 2 ∫0 dx 2 dx 2 dx dx 0 0 = Πe +

l

l

l

l

l

1 d 2w d 2w d 2w 1 d 2 we + − − ( EI ) dx ( EI ) ( EI )dx − ∫ w qdx ∫0 2 ∫0 dx 2 dx 2 EI dx 2 dx 2 0 l

1 d 2w d 2w MM e = Π + ∫ ( 2 EI )dx + ∫ dx − ∫ w qdx 2 0 dx EI dx 2 0 0 e

l

= Πe +

1 d 2w d 2w ( EI )dx ≥ Π e for all virtual w 2 2 ∫0 dx 2 dx

Since the equation in the box represents the total virtual work in a beam, the total potential energy

of a beam becomes minimum for all virtual displacement fields when the principle of virtual work holds. In the above equation, the equality sign holds if and only if w = 0 .


50


Principle of Virtual Work If a beam is in equilibrium, the principle of the virtual work holds for the beam,. l

∫ w ( EI 0

d 4 we − q )dx = 0 for all virtual displacement w dx 4

l

l

l

l

d 2w d 2 we dw d 2 we d 3 we − + − EI dx w qdx EI w EI ∫0 dx 2 dx 2 ∫0 dx dx 2 0 dx 3 l

l

l

=0 0

l

d 2w d 2 we MM e EI dx w qdx − = ∫0 dx 2 dx 2 ∫0 ∫0 EI dx − ∫0 wqdx = 0

In case that there is no support settlement, the boundary terms in above equation vanishes identically since either virtual displacement including virtual rotational angle or corresponding forces (moment and shear) vanish at supports. The principle of virtual work x in a beam by applying an unit load at ~ x yields the displacement of an arbitrary point ~ and by using the reciprocal theorem. l

l

l

l

e

MM ~ ~ ∫0 w qdx = ∫0 wq dx = ∫0 wδ( x − x )dx = w( x ) = ∫0 EI dx z

Approximation using the principle of minimum potential energy -

Approximation of displacement field n

w = ∑ ai g i i =1

-

Total potential energy by the assumed displacement field l

l

l

l

n n n 1 d 2w d 2w 1 Π = ∫ ( 2 EI 2 )dx − ∫ wqdx = ∫ (∑ ai g i′′)EI (∑ a j g ′j′ )dx − ∫ ∑ ai g i qdx 2 0 dx 2 0 i =1 dx j =1 0 0 i =1 h

-

The first-order Necessary Condition n n n ∂Π h ∂ 1 ( ∫ (∑ ai g i′′)EI (∑ ai g i′′)dx − ∫ ∑ ai g i qdx) = ∂a k ∂a k 2 0 i =1 i =1 0 i =1 l

l

l

l

l

n n 1 = [ ∫ g k′′ EI (∑ a j g ′j′ )dx + ∫ (∑ ai g i′′)EIg k′′dx] − ∫ g k qdx) 2 0 i =1 0 i =1 0 l

l

n

i =1 0

0

i =1

n

or

Ka = f

= ∑ ∫ g k′′ EIg i′′dxai − ∫ g k qdx = ∑ K ki ai − f k = 0


51


Example P

i) with one unknown

w = ax( x − l ) = a ( x 2 − lx) → w′′ = 2a l

Π total =

l

l

l l2 1 l2 1 1 2 2 2 ′ ′ EI w dx P x wdx EI a dx aP EI a l aP ( ) ( ) ( 2 ) 4 − δ − = + = + ∫0 2 ∫0 2 2 ∫0 4 2 4

∂Π total l2 Pl Pl = 0 → 4aEIl + P = 0 → a = − →w=− ( x 2 − xl ) ∂a 4 16 EI 16 EI l Pl 3 l Pl 3 Pl 3 w( ) = , we ( ) = = 0.0208 , Error = 2 48 EI 2 64 EI EI

l l w e ( ) − w( ) 2 2 = 0.25 l we ( ) 2

ii) with two unknowns

w = ax( x − l ) + bx( x 2 − l 2 ) → w′′ = 2a + 6bx l

Π total = =

l

l

1 1 3l 3 l l2 2 2 ′ ′ − δ − = + − − − ) ( ) ( ) ( 2 6 ) ( EI w dx P x wdx EI a bx dx P a b ∫0 2 ∫0 2 2 ∫0 4 8 1 3l 3 l2 l3 l2 ) EI (4a 2 l + 24ab + 36b 2 ) + P(a + b 2 2 3 4 8

∂Π total l2 = 0 → EI (4la + 6l 2 b) = − P ∂a 4 ∂Π total 3l 3 = 0 → EI (6l 2 a + 12l 3 b) = − P ∂b 8

→a=−

1 Pl , b = 0 ??? 16 EI

iii) with three unknowns

w = ax( x − l ) + bx( x 2 − l 2 ) + cx( x 3 − l 3 ) → w′′ = 2a + 6bx + 12cx 2 l

Π total =

l

1 l EI ( w′′) 2 dx − ∫ Pδ( x − ) wdx ∫ 20 2 0 l

l2 1 3l 3 7l 4 2 2 = ∫ EI (2a + 6bx + 12cx ) dx − P(−a − b −c ) 20 4 8 16 l3 l5 l2 l3 l4 1 EI (4a 2 l + 36b 2 + 144c 2 + 24ab + 48ac + 144bc ) + 2 3 5 2 3 4 2 3 4 3l 7l l P(a + b +c ) 4 8 16

=


52

School of Civil, Urban & Geosystem Eng., SNU ∂Π total l2 2 3 = 0 → EI (4la + 6l b + 8l c) = − P 4 ∂a ∂Π total 3l 3 = 0 → EI (6l 2 a + 12l 3 b + 18l 4 c) = − P 8 ∂b ∂Π total 144 5 7l 4 l c) = − P = 0 → EI (8l 3 a + 18l 4 b + 5 16 ∂c

w=

1 Pl  a = 64 EI  5 P  → b = − 32 EI  5 P  c = 64 EIl 

5 P 5 P 1 Pl x( x − l ) − x( x 2 − l 2 ) + x( x 3 − l 3 ) 64 EI 32 EI 64 EIl

1 1 5 3 5 7 21 Pl 3 l Pl 3 Pl 3 (− + − )= = 0.0205 , Error = 0.0144 w( ) = 2 64 4 32 8 64 16 1024 EI EI EI iv) with one sin function w = a sin

l π π 1 1 = ∫ EI ( w′′) 2 dx − ∫ Pδ( x − ) wdx = ∫ EI (a( ) 2 sin x) 2 dx − aP l l 20 2 20 0 l

Π total

π π π x =→ w′′ = a( ) 2 sin x l l l

l

l

π π π l 1 1 EIa 2 ( ) 4 ∫ sin 2 xdx + aP = EIa 2 ( ) 4 + aP l 0 l l 2 2 2 l

=

∂Π total π l π 2 Pl 3 2 Pl 3 = 0 → EIa( ) 4 − P = 0 → a = 4 →w= 4 sin x ∂a l l 2 π EI π EI l 1 Pl 3 l Pl 3 , w( ) = , we ( ) = 2 48.7045 EI 2 48EI

Error = 0.0145

v) with two sin function w = a sin

π 3π 3π 3π π π x + b sin x =→ w′′ = a( ) 2 sin x + b( ) 2 sin x l l l l l l

l

Π total

l

1 l = ∫ EI ( w′′) 2 dx − ∫ Pδ( x − ) wdx 20 2 0 1 3π 3π 2 π π EI (a( ) 2 sin x + b( ) 2 sin x) dx − aP + bP ∫ 20 l l l l l

=

1 3π π π π 3π π xdx + = EIa 2 ( ) 4 ∫ sin 2 xdx +EIab( ) 2 ( ) 2 ∫ sin sin 2 l 0 l l l 0 l l l

l

1 3π 3π EIb 2 ( ) 4 ∫ sin 2 xdx − aP + bP 2 l 0 l l

=

1 π l 1 3π l EIa 2 ( ) 4 + EIb 2 ( ) 4 − aP + bP 2 l 2 2 l 2


53

School of Civil, Urban & Geosystem Eng., SNU ∂Π total π l 2 Pl 3 = 0 → EIa( ) 4 − P = 0 → a = 4 ∂a l 2 π EI ∂Π total 3π l 2 Pl 3 = 0 → EIb( ) 4 + P = 0 → b = − ∂b l 2 (3π) 4 EI π 2 Pl 3 1 3π (sin x − sin x) 4 l 81 l π EI l Pl 3 Pl 3 l Pl 3 w( ) = 0.0205(1 + 0.0123) = 0.0208 , w e ( ) = 0.0208 , Error ≅0 2 2 EI EI EI w=

4.3 Truss problems − F ji − Fmi ( i )

X

i

Xi

Y

i

− F1i

Yi z

Potential Energy Π int =

njn 1 nmb ( Fi ) 2 li Π = − , ( X i u i + Yi vi ) ∑ ∑ ext 2 i =1 EA i i =1

Π total = Π int + Π ext =

1 nmb ( Fi ) 2 li njn − ∑ ( X i u i + Yi vi ) ∑ 2 i =1 EA i i =1

where nmb and njn denotes the total number of members and the total numbers of joints in a truss. z

Equilibrium Equations m(i )

m(i )

− ∑ H + X = 0 , − ∑ V ji + Y i = 0 j =1

i j

i

j =1

for

i = 1, L , njn

where m(i), H ij and V ji are the number of member connected to joint i, the horizontal component and the vertical component of the bar force of j-th member connected to joint i, respectively.


54


Principle of Minimum Potential Energy Π total

1 nmb ( Fi e + Fi ) 2 l i njn = ∑ − ∑ ( X i (u i +u i ) + Yi (vi + vi )) 2 i =1 EAi i =1 =

1 nmb ( Fi e ) 2 li njn 1 nmb ( Fi ) 2 li nmb Fi e Fi li njn ( ) X u Y v − + + +∑ − ∑ ( X i u i + Yi vi ) ∑ ∑ ∑ i i i i 2 i =1 EAi 2 i =1 EAi EAi i =1 i =1 i =1

=

nmb ( Fi ) 2 li 1 nmb ( Fi e ) 2 li njn ( ) X u Y v − + + ∑ ∑ ∑ i i i i 2 i =1 EAi EAi i =1 i =1

1 nmb ( Fi ) 2 l i =Π + ∑ ≥ Πe 2 i =1 EAi e

for all virtual displacement fields

where Fi e and Fi are the bar force of i-th member induced by the real displacement of joints and virtual displacement induced by the virtual displacement of joints. Since the equation in the box represents the total virtual work in a truss, the total potential energy of a truss becomes

minimum for all virtual displacement fields when the principle of virtual work holds. In the above equation, the equality sign holds if and only if the virtual displacements at all joints are zero. z

Virtual Work Expression If a truss is in equilibrium, the principle of the virtual work holds for the truss,. njn

m(i )

m(i )

i =1

j =1

j =1

∑ ((− ∑ H ij + X i )u i + (-∑V ji + Y i )v i ) = 0

(v 2 − v1 ) sin θ i (v 2 − v1 )

(u 2 − u1 ) cos θ i

(u 2 − u1 )

θi

θi


55

School of Civil, Urban & Geosystem Eng., SNU njn

m(i )

m(i )

i =1

j =1

j =1

∑ ((− ∑ F ji cos θ j + X i )u i + (-∑ F ji sin θ j + Yi )vi ) = 0 m(i )

njn

∑ (u ∑ F i

i =1

j =1

i j

m(i )

njn

j =1

i =1

cos θ j + δv i ∑ F ji sin θ j ) = ∑ ( X i u i + Yi vi )

nmb

njn

i =1

i =1

∑ ( Fi e cos θ i (ui2 − u i1 ) + Fi e sin θi (vi2 − vi1 )) = ∑ ( X i ui + Yi vi ) nmb

nmb

nmb

i =1

i =1

∑ Fi e (cos θ i (ui2 − u i1 ) + sin θ i (vi2 − vi1 )) = ∑ Fi e ∆li = ∑ Fi e i =1

nmb

∑ i =1

nmb Fi l i F eF l =∑ i i i ( EAi ) i =1 ( EAi )

e

n Fi Fi l i = ∑ ( X i u i + Yi vi ) ( EAi ) i =1

The principle of virtual work yields the displacement of a joint k in a truss by applying an unit load at a joint k in an arbitrary direction and by using the reciprocal theorem. Fi e Fi l i X k u k + Yk v k = X u k cos α = u k cos α = ∑ i =1 ( EAi ) nmb

Since α represnts the angle between the applied unit load and the displacement vector, u k cos α are the displacement of the joint k in the direction of the applied unit load.



56

Chapter 5 Matrix Structural Analysis

Mr. Force & Ms. Displacement Matchmaker: Stiffness Matrix


57

School of Civil, Urban & Geosystem Eng., SNU 5.1 Truss Problems 5.1.1 Member Stiffness Matrix R f yR δ y

f yL δ Ly

f xL δ Lx

z

f xR δ Rx

Force – Displacement relation at Member ends EA R (δ x − δ Lx ) L EA R (δ x − δ Lx ) f xR = L L R fy = fy = 0

f xL = −

z

Member Stiffness Matrix in Local Coordinate System       

e

f xL   1   L fy  EA  0 = f xR  L − 1  R fy   0

0 − 1 0 δ Lx    0 0 0 δ Ly    0 1 0 δ Rx   0 0 0 δ Ry 

e

(f )e = [k ]e (δ )e z

Transformation Matrix

Vy

vy

vx

②

Vx

φ

① V x = cos φv x − sin φv y  V x → V y = sin φv x + cos φv y  V y

 cos φ − sin φ v x  =  v   sin cos φ φ   y  

(V ) = [ γ ]T ( v) → ( v) = [ γ ](V )


58


Member End Force e

 Fx1  0  cos φ − sin φ 0  1   sin φ cos φ 0 0   Fy    F2  =  0 0 cos φ − sin φ  x     F2  φ φ 0 0 sin cos    y  z

e

f xL   f yL  → (F)e = [Γ]T (f )e f xR   f yR 

Member End Displacement e

e

 δ Lx  0 0  ∆1x   cos φ sin φ  L    − sin φ cos φ 0 0  ∆1y  δy    2 → (δ)e = [Γ](∆)e δR  =  0 φ φ 0 cos sin  ∆ x   x   2  δR  − φ φ 0 0 sin cos   ∆ y   y z

Member Stiffness Matrix in Global Coordinate (F)e = [Γ]T (f )e = [Γ]T [k ]e (δ ) e = [Γ]T [k ]e [Γ](∆ ) e (F ) e = [K ] e ( ∆ ) e

− cos 2 φ − sin φ cos φ cos 2 φ sin θ cos φ    − sin 2 φ [K 11 ]e [K 12 ]e  sin 2 φ − sin φ cos φ EA  sin θ cos φ e [K ] = =  L  − cos 2 φ − sin θ cos φ cos 2 φ sin φ cos φ [K 21 ]e [K 22 ]e    − sin 2 φ sin φ cos φ sin 2 φ − sin φ cos φ

5.1.2 Global Stiffness Equation

− (F1 ) m (3) z

Nodal Equilibrium

− (F 2 ) m ( 2 )

n-th joint

− (F 2 ) m ( 4 )

− (F1 ) m (1)

− (F1 ) m ( 5) Pm

m-th joint

Pm

i-th member (P) m = (F 1 ) m (1) + (F 2 ) m ( 2) + (F 1 ) m ( 3) + (F 2 ) m ( 4) + (F 1 ) m (5) =

nm ( m )

∑ (F

1 or 2 m ( k )

)

k =1


59


Pn

(F 2 ) i

n-th joint 2

1 i-th member

Pm m-th joint

 0  0  nm (1) 1 or 2 1( k )     ∑ (F )  M M   (P)   k =1      1 i   M   ( ) F I M   p  p nm ( m )     1 or 2 ( ) m k m ( P ) =  ( P )  =  ∑ (F )  = ∑  M  = ∑  M   k =1  i =1  (F 2 )i  i =1 0  M   M    nm ( q )    q   M  1 or 2 q ( k )   (P)   M ( ) F  ∑    0  k =1   0  1

(F1 )i

0 M  m-th row  0 1 i p  (F )  M  2 i  = ∑ [E]i (F)i = [E](F )  (F )  i =1 I  n-th row  M 0

 (F ) 1     M  [E] = [[E]1 L [E]i L [E] p ] , (F) =  (F) i     M   p   (F ) 


60


Compatability Condition un

n

(∆ 2 )i

2

m

um

1

(∆ 1 )i

i-th member  ( ∆ 1 )i   I 0  u m    (∆ 1 )i = u m , ( ∆ 2 )i = u n → ( ∆ )i =  2 i  =   (∆ )  0 I  u   n    u1     M  u  m 1 i   (∆ ) 0 L I L 0 L 0    i    M  = [C]i (u) (∆ ) = =  ( ∆ 2 ) i  0 L 0 L I L 0     u     n  M  n-th column  u   q

m-th column

 (∆ )1  [C]1  u1       (∆ ) =  M  =  M  M  = [C](u)  (∆ ) p  [C] p  u      q  Compatibility Matrix z

Contragradient (P)T ⋅ (u) = (F)T ⋅ (∆) → (P)T ⋅ (u) = (F)T [C](u) → ((P)T − (F)T [C])(u) = 0 for all possible (u) → (P)T = (F)T [C] (P) = [C]T (F) = [E](F) → [C]T = [E]


61


Unassembled Member Stiffness Equation  (F )1  [K ]1     M   M  (F ) i  =  0     M   M  p   (F)   0

z

L

0

L

L

M

L

L [K ] i L L

M

L

L

0

L

0  ( ∆ )1    M    i   → (F) = [K ](∆) 0  (∆)   M  M    [K ] p  (∆) p 

Global Stiffness Equation (P) = [C]T (F) = [C]T [K ](∆) = [C]T [K ][C](u) (P) = [K ](u)

z

where

[K ] = [C]T [K ][C]

Direct Stiffness Method

[

[K ] = [C]T [K ][C] = [C]1 L[C]i L[C] p T

T

T

]

[K ]1   M  0   M  0 

L L

0 M

L [K ] i L M L 0

1 0  [C]     M  i L 0  [C]    L M  M  p L [K ] p  [C] 

L L

= [C]1 [K ]1 [C]1 + L + [C]i [K ]i [C]i + L + [C] p [K ] p [C] p T

T

0 M  I  iT i i [C] [K ] [C] =  M M  M 0  0 M  I  = M M  M 0 

0 M  0 i  [K ] M   11 i [K ] I   21  M 0

T

[K 12 ]i  0 L I L 0 L 0   [K 22 ]i  0 L 0 L I L 0

0 M  0 i  0 L [K 11 ] M  0 L [K 21 ]i   I  M 0

m-th row

0 0 0 [ K ] i 11 L [K 12 ]i L 0  0  = 0 L [K 22 ]i L 0  0 [K 21 ]i  n-th row 0 0 m-th column

0 0 0 0 0

0 [K 12 ] 0 0 0  i [K 22 ] 0 0 0 n-th column 0

i


62

School of Civil, Urban & Geosystem Eng., SNU 5.1.3 Example

P4, u4

P3, u3 2 L

2

1

P2, u2 1 P1, u1

P6, u6

3

P5, u5

3 2L

z

Member Stiffness Matrix − cos 2 φ − sin φ cos φ cos 2 φ sin φ cos φ    − sin 2 φ [K 11 ]e [K 12 ]e  sin 2 φ − sin φ cos φ EA  sin φ cos φ = [K ] e =  L  − cos 2 φ − sin φ cos φ cos 2 φ sin φ cos φ [K 21 ]e [K 22 ]e    − sin 2 φ sin φ cos φ sin 2 φ − sin φ cos φ

- Member 1: φ = 45o  1  2  1 EA  2 [K ]1 = 2 L − 1  2  1 −  2

1 2 1 2 1 − 2 1 − 2

−

1 2 1 − 2 1 2 1 2

1 −  2 1 −  2 1 2 1  2

 1  2  1 EA − 2 [K ] 2 = 2 L − 1  2  1   2

1 2 1 2 1 2 1 − 2

1 2 1 2 1 2 1 − 2

1 2 1 −  2 1 − 2 1  2

- Member 2: φ =-45 o −

−


63

School of Civil, Urban & Geosystem Eng., SNU - Member 3: φ = 0o  1  EA  0 [K ] 3 = 2 L − 1   0 z

Equilibrium Equation

0 − 1 0 0 0 0 0 1 0  0 0 0

P4 P3

P2

P6 P5

P1

P1 = ( Fx1 )1 + ( Fx2 ) 3 P2 = ( F ) + ( F ) 1 1 y

2 3 y

P3 = ( Fx2 )1 + ( Fx1 ) 2 P4 = ( Fy2 )1 + ( Fy1 ) 2 P5 = ( Fx2 ) 2 + ( Fx1 ) 3 P6 = ( Fy2 ) 2 + ( Fy1 ) 3

 P1  1     P2  0  P  0  3=  P4  0     P5  0  P  0  6 

0 1 0 0 0 0

0 0 1 0 0 0

[E]1

0 0 0 1 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

[E]2

0 0 0 0 0 1

0 0 0 0 1 0

0 0 0 0 0 1

1 0 0 0 0 0

[E]3

  F 1 1   x     Fy1    2     Fx    2   0   Fy   2 1  Fx1      0  Fy1      0  Fx2     0  Fy2     0  F 1  3  x    F1     y2     Fx    F 2    y    

(P) = [[E1 ] [E 2 ] [E 2 ]](F) = [E](F)


64


Compatibility Condition u4 u3

u2

u6

u5

u1   ∆1 1    x   1   ∆1y      2   0   ∆ x   0   ∆2      y  2  0   ∆1x   0  1      ∆ y   0   ∆2   = 0   2x      ∆ y   0  1 3    ∆ x   0   ∆1   0   2y      ∆ x   1   ∆2   0  y    

0 1 0 0 0 0 0 0 0 0 0 1

0 0 1 0 1 0 0 0 0 0 0 0

0 0 0 1 0 1 0 0 0 0 0 0

0 0 0 0 0 0 1 0 1 0 0 0

0 [C]1 0 0  0  u1   u1      0  u 2  1  u2   [C]   0  u 3   2   u 3    = [C]   0  u 4   3  u 4     [C]    u 1  u 5    5    u  0  u6   6  1 [C]2  0 [C]3 0

(∆ ) = [C](u) [E1 ] = [C1 ]T , [E 2 ] = [C 2 ]T , [E 3 ] = [C 3 ]T → [E] = [C]T z

Global Stiffness Matrix

[K ] = [C]T [K ][C] = [C]1 [K ]1 [C]1 + [C] 2 [K ] 2 [C] 2 + L + [C]3 [K ]3 [C]3 T

T

T


65

School of Civil, Urban & Geosystem Eng., SNU 1 0  EA 0 [C1 ]T [K 1 ][C1 ] =  2 L 0 0  0 1 0  EA 0 =  2 L 0 0  0

0 1 0 0 0 0

0 0 1 0 0 0

0 1 0 0 0 0

0  1  0  12 0  2  1 − 1  0  2  1 0 −  2

1 2 1 2 1 − 2 1 − 2

0 0  1 EA [C 2 ]T [K 2 ][C 2 ] =  2 L 0 0  0 1 0  EA 0 =  2 L 0 0  0

0 1 0 0 0 0

0 0 1 0 0 0

0  0  0  0  0  1  0  0   0 0 

0 0  EA 0 [C 3 ]T [K 3 ][C 3 ] =  2 L 0 1  0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

1 2 1 0 − 2 1 0 − 2 1 0 2

0

0 0 0 0 0 1

1 0 0 0 0 0

0  1  0  12 0  2  1 − 1  0  2  1 0 −  2 1 2 1 − 2 1 2 1 2 −

1 2 1 − 2 1 2 1 2 −

0  1  0  12 0  − 2  0  − 1  0  2  1 1   2 1 2 1 2 1 2 1 − 2

−

0 1  1 0  0  0  − 1  0  0  0

1 −  2 1  1 0 −  0 1 2  1  0 0 2  0 0 1  2  1  2  0 0  1    2 0 0 EA − 1 = 2 L  2 0 0 1  −   2 0 0  0   0

1 2 1 2 1 − 2 1 − 2

1 2  0 1 −  0 2  1  0 −  2  0 1  2 0 0 1  2 0 1  −  2  = EA 0 1 2L  −    2 0  1   2 0 

1 2 1 2 1 2 1 − 2

−

1 2 1 2 1 2 1 − 2

−

1 2 1 − 2 1 2 1 2 −

1 2 1 2 1 2 1 − 2

0 0 1 0

0 0 0 1

0 0 0 0

0 0 0  0

1 1 1 − − 2 2 2 1 1 1 − − 2 2 2 1 1 1 − 2 2 2 1 1 1 − 2 2 2 0 0 0 0 0 0

 0 0  0 0  0 0   0 0  0 0 0 0

−

0 − 1 0 0 0 0 0 0 0 1 0  1  0 0 0 0

0 0 0 1

0 0 0 0

0 0 0 0

0 0 0 0 0 0

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0  1

0 0 0 0 0 0 0 0 1 1 1 1  − − 2 2 2 2 1 1 1 1 − −  2 2 2 2 1 1 1 1 − −  2 2 2 2 1 1 1 1 − − 2 2 2 2 

1 0 0 0

0 1 0  0


66

School of Civil, Urban & Geosystem Eng., SNU 0 0  EA 0 =  2 L 0 1  0

0 0 0 0 0 1

1 0 0 0 0 0

0 1 − 1 0  0  0  1  0  0  0

1  1  2 2  1 1  2  2 1 EA  1 − − [K ] = 2 2L  2  1 1 − − 2  2 0 0   0 0  1 0 0  0 0 0  EA  0 0 0 +  2L  0 0 0 − 1 0 0   0 0 0 1  1 2 2 + 2  1   2 2  1 −  EA  2 2 = L  − 1  2 2  1  − 2   0  

0 0 0 0

0 0 0 0

 1  0 1 0  0 0 EA  0 =  − 1 0 2 L  0  − 1 0 0   0

0 0 0 0

0 0 0 0 0 0

0 − 1 0 0 0 0 0 0 0  0 0 0 0 1 0  0 0 0

0 0 0 0 0 0

 0 0 0 0 0 0 0 0 0 0 0 0 0 0   1 1 1 1 0 0 0 0  − −  2 2 2 2  EA  1 1 1 1 0 0 + 0 0 − −  2L  2 2 2 2    1 1 1 1 0 0 −  0 0 − 2 2 2 2   0 0 1 1 1 1 0 0 − −  0 0 0 2 2 2 2  0 0 0  0 0  0

1 1 − 2 2 1 1 − − 2 2 1 1 2 2 1 1 2 2 0 0 −

0 0 −1 0 0 0

0

0

0

0

1

0

0 1

2 2 1 2 2 1 − 2 2 1 − 2 2 0 0

− − 1 2 2 1 2 2 −

1

−

2 2 1

−

2 2 + −

1 2 2 1

2 2 1

2 2 1

2 2

1

−

2 2 1

2 2 1 1 − 2 2 2 2 1 1 + 2 2 2 2 1 2 2 1 − 2 2

1 2

0 −

1 2 2 1

2 2 1 1 + 2 2 2 1 − 2 2

   0   1   2 2  1  − 2 2 1   − 2 2 1   2 2  0


67


Stiffness Equation

1 + 2   2 2  1 Unknown  P1   2 2    1 P Unknown  2  − Known  P3  EA  2 2  = L − 1 Known  P4     Known  P5   2 2  −1 Unknown  P6   2   0  z

2 2 1 2 2 1 − 2 2 1 − 2 2 0 0

− −

1 2 2 1 2 2 1

− −

−

1 2 2 1

2 1

2 2 1+ 2

2 2 1 − 2 2

2 2 1 − 2 2

1

2 2

1 2

0

2 2

1

0

2 2 1

−

2 2 1 0

2

−

1

   0  u1    1  u 2    2 2  u 3  1  u 4  −  2 2  u 5    1  u  −  6 2 2 1  2 2  0

Known Known Unknown Unknown Unknown Known

Application of Support Conditions (Boundary Conditions) 1 + 2   2 2  1  P1   2 2    1  P2  − P  EA 3  2 2  =  P4  L − 1     2 2  P5  P   −1  6  2   0 

z

1

1 2 2 1 2 2 1 − 2 2 1 − 2 2 0 0

− −

1 2 2 1 2 2 1 2 0

−

1 2 2 1

2 2

− −

1

−

2 2 1

0

2 2 0 1

1 2

−

1 2 2 1

2 1

2 2 1+ 2

2 2 1 − 2 2

2 2 1 − 2 2

   0  u1    1  u 2    2 2  u 3  1  u 4  −  2 2  u 5    1  u  −  6 2 2 1  2 2  0

Final Stiffness Equation

 1   P3   2   EA   P4  =  0 L P    5 − 1  2 2

0 1 2 1 2 2

 1 1  −   2 2  u 3   u3   2   L  1    u4  →  u4  =  0 EA 2 2   u   u   5 1 + 2  5  − 1  2 2 2 2 

0 1 2 1 2 2

1  −  2 2 1  2 2   1+ 2  2 2 

−1

 P3     P4  P   5


68

School of Civil, Urban & Geosystem Eng., SNU 5.2 Beam Problems 5.2.1 Member Stiffness Matrix

R f yR δ y

f yL δ Ly

m R θR

m L θL

z

Force-Displacement Relation at Member Ends

mL =

4 EI e L 2 EI e R 6 EI e L 6 EI e R θ + θ + 2 δy − 2 δy Le Le Le Le

mR =

2 EI e L 4 EI e R 6 EI e L 6 EI e R θ + θ + 2 δy − 2 δy Le Le Le Le

M eL + M eR 6 EI e L 6 EI e R 12 EI e L 12 EI e R f = = 2 θ + 2 θ + 3 δy − 3 δy Le Le Le Le Le L y

6 EI 6 EI 12 EI 12 EI M eL + M eR = − 2 e θ L − 2 e θ R − 3 e δ Ly + 3 e δ Ry f =− Le Le Le Le Le R y

z

Transformation Matrix is not required

f →F, m→M z

,

δ→∆

,θ → Θ

Member Stiffness Matrix e

 Fy1   12    L2    e  1  6 M  EI e  Le     = L  12 e −  Fy2  2  Le    6     2   Le M  

e

6 Le

12 − 2 Le 6 4 − Le 6 12 − Le L2e 6 2 − Le

1 6  ∆ y  Le     1 2 Θ    or (F)e = [K ]e (∆ )e    6 −  ∆2  Le  y    4   Θ 2y 


69

School of Civil, Urban & Geosystem Eng., SNU 5.1.2 Global Stiffness Matrix z

Nodal Equilibrium

Vi Mi

i-th Member

(i-1)-th Member

 → P i = (F) iR−1 + (F) iL 2 1  i M = ( M ) i −1 + ( M ) i 

V i = ( Fy2 ) i −1 + ( Fy1 ) i

  P 1   (F )1L    R L 2  P  (F ) 1 + ( F ) 2      M M      P i −1   (F ) R + (F) L  i −1     i−2 i R L  P  =  (F ) i −1 + (F) i  = [E]1 L [E]i L [E] p   i +1   R L  P   (F ) i + (F) i +1   M   M     p R L   P  (F ) p −1 + (F) p   p +1   R   P ( ) F    p 

[

i-th row (i+1)-th row

z

Compatibility

 0  0     M  M  (F ) L   I  iR  =   ( F ) i  0  M  M    0  0   

]

0 M  0 (F )iL   = [E]i (F)i  I  (F)iR  M  0 u 2 i +1

u2i −1 u2i

(i-1)-th Member

 (F ) 1     M   (F)  = [E](F) i    M   (F )  p  

u2i + 2

i-th Member

(∆1y )i = u2i −1 (Θ1 )i = u2i (∆2y )i = u2i +1

→

 (∆ ) iL   I 0 u i  = (∆ ) iL = u i , (∆ ) iR = u i +1 → (∆ ) i =   i +1  R  0 I ( ) ∆   u  i  

(Θ 2 ) i = u 2 i + 2


70

School of Civil, Urban & Geosystem Eng., SNU i-th column

 u1    M   (∆ ) iL  0 L I 0 L 0 u i    = (∆ ) i =   i +1  = [C]i (u) R  I L L 0 0 0 ∆ ( ) u   i     M   (i+1)-th column  p +1  u   (∆ )1  [C]1  u1       (∆ ) =  M  =  M M  = [C](u)  (∆ )  [C]  u p +1  p    p   z

Unassembled Member Stiffness Equation

 (F)1  [K ]1     M   M     (F ) i  =  0  M   M     (F ) p   0 z

L

0

L M L [K ] i L M L 0

L L L L L

0  ( ∆ )1    M     0  (∆) i  →  M  M    [K ] p  (∆) p 

(F) = [K ](∆)

Global Stiffness Equation (P) = [C]T (F) = [C]T [K ](∆) = [C]T [K ][C](u) (P) = [K ](u)

z

where

[K ] = [C]T [K ][C]

Direct Stiffness Method 0 0 0  M M M 0 [ K L  11 ]i [C]Ti [K ]i [C]i =  0 L [K 21 ]i  M M M  0 0 0

0

M [K 12 ]i [K 22 ]i M 0

i-th column

0  M M L 0 L 0  M M  0 0 0

i-th row i+1-th row

i+1-th column


71

School of Civil, Urban & Geosystem Eng., SNU 5.2.3 Example

V 1 , w1 M 1 , θ1

z

V 3 , w3

V 2 , w2

V 4 , w4 M 3 , θ3

M 2 , θ2

M 4 , θ4

Equilibrium Equation V 1 = V1L

, M 1 = M 1L

V 2 = V1R + V2L , M 2 = M 1R + M 2L V 3 = V2R + V3L , M 3 = M 2R + M 3L V 4 = V4R

V 1  M 1  2 V M 2 (P) =  3 V  3 M  4 V M 4 

 1    0    0  0 =  0    0  0    0 

0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0 [E]1

z

0 0 0 1 0 0 0 0

, M 4 = M 4R

0 0 1 0 0 0 0 0

0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0

[E]2

0 0 0 0 0 1 0 0

0 0 0 0 1 0 0 0

0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0 [E]3

V1L     M 1L   R  0V1   R 0 M 1  L 0  V2     0 M 2L   = [E](F)  0V2R   0  M R   2  0  V L   3 1 L  M3   R  V3  M R   3 

Compatability Condition

w1L = w1 , θ1L = θ1 , w1R = w 2 , θ1R = θ 2 w2L = w 2 , θ 2L = θ 2 , w2R = w 3 , θ 2R = θ 3 w3L = w 3 , θ 3L = θ 3 , w3R = w 4 , θ 3R = θ 4


72

School of Civil, Urban & Geosystem Eng., SNU  w1L   1   θ1L    R  0  w1  0  R    θ1  0  w2L  0  L    θ2  0 (∆) =  R  =   w2   0  θ R  0  2    w3L  0  L  0  θ3    R  0  w3  0  θR    3  z

0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 1 0 1 0 0 0 0 0 0

0 0 0 0 0 0 1 0 1 0 0 0

0 0 0 0 0 0 0 1 0 1 0 0

0 0 0 0 0 0 0 0 0 0 1 0

0 0 1 0 w  1 0  θ  2 0  w  0 θ 2 0  w 3  0  θ 3 0  w 4  0  4  θ 0 1

[C]1  w1     θ1   2  w   [C]1  θ 2  = [C] 2   w 3    [ ] C   3  3 θ   4  w   θ4    [C]3 [C]

       = [C](u)       

2

Unassembled Member Stiffness Matrix 0  (F)1  [K ]1    (F ) =  (F ) 2  =  0 [K ] 2  (F )   0 0  3 

z

0 0 1 0 1 0 0 0 0 0 0 0

0   (∆) 1    0  (∆) 2  = [K ](∆) [K ]3  (∆) 3 

Global Stiffness Equation (P) = [E](F) = [E][K ](u) = [C]T [K ][C](u) = [K ](u) 0 0  [C]1  [K ]1 (P) = [C]1T , [C]T2 , [C]T3  0 [K ] 2 0  [C] 2  (u)  0 0 [K ]3  [C]3  = [K ](u) = ( [C]1T [K ]1 [C]1 + [C]T2 [K ] 2 [C] 2 + [C]T3 [K ]3 [C] 3 )(u)

[

]


73

School of Civil, Urban & Geosystem Eng., SNU -

Member 1 1 0 0  EI 1 0 L1 0 0 0  0

-

0 1 0 0 0 0 0 0

 12  L2  1  6  L1  12  EI 1 − L2 1 L1  6   L1  0  0  0   0 Member 2 0 0 1  EI 2 0 L2  0 0 0  0

0 0 0 1 0 0 0 0

0 0  0   0 EI 2   L2  0   0  0 0

0 0 0 0 0 0 0 0

12 0  2  0  L 0  6  1  L1 0  12 − 0  L2 1 0  6  0  L  1 6 12 − 2 L1 L1 6 4 − L1 6 12 − L1 L12 6 2 − L1 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0

0  12  2 0   L2 0  6  0   L2 0  12 − 1  L22 0  6  0  L  2 0 0 0 0 12 6 L22 L2 6 4 L2 12 6 − 2 − L2 L2 6 2 L2 0 0 0 0

0 0 0 0 1 0 0 0

6 L1

12 L12 6 4 − L1 6 12 − L1 L12 6 2 − L1 6 L1 2 6 L1 4 0 0 0 0 6 L2

−

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

0 0 0 0

0 0 0 0

0 0 = 0  0

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

0 0 = 0  0

 0 0 0 0  0 0 0 0   0 0 0 0   0 0 0 0  0 0 0 0 0 0 0 0 0 0 0 0  0 0 0 0

12 L22 6 4 − L2 6 12 − L2 L22 6 2 − L2 0 0 12 − 2 L2 6 − L2 12 L22 6 − L2 0 0

6 L1   2  1  0 6  0  − L1  0  4 

−

0 0 6 L2 2 −

6 L2 4 0 0

6 L2   2  0  0 6  0 −  L2   0  4  0 0 0 0  0 0   0 0  0 0   0 0  0 0 0 0


74

School of Civil, Urban & Geosystem Eng., SNU -

z

Member 3 0 0 0  EI 3 0 L3 1 0 0  0

0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0

0  12  2 0  L3 0  6  0  L3 0  12 − 0  L23 0  6  1  L  3

0 0 0  0 0 EI 3   L3 0   0  0  

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 12 L23 6 L3 12 − 2 L3 6 L3

6 L3

12 L23 6 4 − L3 6 12 − L3 L23 6 2 − L3 0 0 0 0 6 L3 4 −

6 L3 2

−

0 0 0 0 12 − 2 L3 6 − L3 12 L23 6 − L3

6 L3   2  0  0 6  0 −  L3  0  4  0  0  0   0  6  L3   2   6 −  L3  4  

0 0 0 0

0 0 0 0

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 = 0  1

Global Stiffness Matrix 12 I 6I1   12 I 1 6 I 1 0 0 0 0  − 31 2 2  L3 L1 L1 L1 1   6 4 6 2 I I I I1 1 1 1  0 0 0 0  − 2   L12 L1 L1 L1   12 I 6 I 12 I 1 12 I 2 6 I 1 6 I 2 12 I 2 6I 2  − 3 1 − 21 0 0 + − + − L1 L13 L32 L12 L22 L32 L22   L1   6I1 2 I1 6I1 6I 2 4 I1 4 I 2 6I 2 2I 2 0 0 − + + −   L12 L1 L12 L22 L2 L2 L12 L2   E 12 I 2 6I 2 12 I 2 12 I 3 6 I 2 6 I 3 12 I 3 6 I 3   0 0 − 3 − 2 + 3 − 2 + 2 − 3  L2 L2 L32 L3 L2 L3 L3 L23    6I 3 2I 3  6I 2 2I 2 6I 2 6I 3 4I 2 4I 3  0 0 − 2 + 2 + − 2  L22 L2 L2 L3 L2 L3 L3 L3   12 I 3 6I 3 12 I 3 6I 3   0 0 0 0 − 3 − 2 − 2  L3 L3 L33 L3    6I 3 2I 3 6I 3 4I 3  0 0 0 0 − 2   2 L L L3 L3  3 3 


75


Application of support Conditions

6I1  12 I 1  L3 L12 1  4I1  6I1 2  L1 L1 V 1   12 I   6 I − 3 1 − 21 M 1  L1  L1  2   6I1 V  2I1  M 2  2 L1   = E  L1 3  V  0 0   3  M    4  0 0  V  M 4      0 0   0 0  

z

12 I 1 6I1  0 0 0 0  3 2 L1 L1  6I1 2I1 − 2 0 0 0 0   w1 L1 L1  12 I 1 12 I 2 6I1 6I 2 12 I 2 6I 2   θ1 + − + − 0 0 3 3 2 2 3 2 L1 L2 L1 L2 L2 L2  2  w 6I1 6I 2 4I1 4I 2 6I 2 2I 2 − 2 + 2 + − 2 0 0  2 L1 L2 L2 L2 L1 L2  θ 6I 12 I 6 I 3  w 3 12 I 6I 12 I 2 12 I 3 6I − 32 − 22 + 3 − 22 + 23 − 3 3 3 L2 L2 L2 L3 L2 L3 L3 L23  θ 3  6I 3 2 I 3  4 6I 2 2I 2 6I 2 6I 3 4I 2 4I 3 − 2 + 2 + − 2 w L22 L2 L2 L3 L2 L3 L3 L3  4  12 I 3 6I 3 12 I 3 6 I 3  θ − 3 − 2 − 2 0 0 L3 L3 L33 L3  6I 3 2I 3 6I 3 4I 3  − 0 0  L23 L3 L23 L3  −

Final Stiffness Equation  4 I1 4I 2 L + L 2  2 M 2  2 I 2    M 3   L2  4  = E V   0 M 4      0  

2I 2 L2 4I 2 4I 3 + L2 L3 6I − 23 L3 2I 3 L3

0 6I 3 L23 12 I 3 L33 6I − 23 L3 −

 0   2 2 I 3  θ 3 L3  θ 6 I  4 − 23  w L3  θ 4 4 I 3   L3 

      


             

76

School of Civil, Urban & Geosystem Eng., SNU 5.3 Frame Problems 5.3.1 Member Stiffness Matrix f yR , δ Ry

f yL , δ Ly

mL , θL

mR , θR

f xL , δ Lx

z

f xR , δ Rx

Force-Displacement Relation at Member Ends - Beam action

mL =

4 EI e L 2 EI e R 6 EI e L 6 EI e R θe + θe + 2 δ y − 2 δ y Le Le Le Le

mR =

2 EI e L 4 EI e R 6 EI e L 6 EI e R θe + θe + 2 δ y − 2 δ y Le Le Le Le

f yL =

M eL + M eR 6 EI e L 6 EI e R 12 EI e L 12 EI e R = 2 θe + 2 θe + 3 δ y − 3 δ y Le Le Le Le Le

f yR = −

6 EI 6 EI 12 EI 12 EI M eL + M eR = − 2 e θeL − 2 e θeR − 3 e δ Ly + 3 e δ Ry Le Le Le Le Le

- Truss action EA R (δ x − δ Lx ) L EA f xR = (δ Rx − δ Lx ) L L R Ve = Ve = 0 f xL = −

z

Member Stiffness Matrix  f xL      0  Ae  L  12 I e  fy   0   L2e    6I e  mL  0   E  Le  =  0 −A  f R  Le  e Ie 12   x  0 −    L2e   R 6I e  0  fy   Le     mR   

0 6I e Le

− Ae

4I e

0

0 6I e − Le

Ae

2I e

0

0

0

0 12 I e − 2 Le 6I − e Le 0 12 I e L2e 6I − e Le

 δ Lx      0   L 6 I e  δ y   Le      θ L  2I e  e     or (f )e = [k ]e (δ )e 0   R 6 I e  δ x  − Le     R 4 I e  δ y      θR   e


77

School of Civil, Urban & Geosystem Eng., SNU Transformation Matrix

z

Vy

vy

vx

②

Vx

θ

① ①②

Vx = cos θvx − sin θv y  V  cos θ − sin θ 0 v    x   x Vy = sin θvx + cos θv y  → Vy  =  sin θ cos θ 0 v y   M   0 0 1 m  M =m     (V ) = [ γ ]T ( v) → ( v) = [ γ ](V ) z

Member End Force e

 Fx1  cos θ − sin θ  1    Fy   sin θ cos θ  1  0 0  M2  =  0  Fx   0 F2   0  y   0 2 M   0 0    z

0 0 0 0 0 0 1 0 0 0 cos θ − sin θ 0 sin θ cos θ 0 0 0

e

0 f xL   L 0 f y  0  m L  e T e  R  → (F) = [Γ] (f ) 0  f x   R 0  f y   1 m R 

Member End Displacement e

e

0 0 0 ∆1x   δ Lx   cos θ sin θ 0  L  ∆1  − sin θ cos θ 0 0 0 0 δy   y    L  0 0 1 0 0 0   θ1  e e  θ Re  =   2  → (δ) = [Γ](∆) ∆ δx  0 0 cos θ sin θ 0 x   0 δR   2  0 0 0 − sin θ cos θ 0 ∆ y   y   θR  0 0 0 0 1 θ1   0  e



78

Member Stiffness Matrix in Global Coordinate (F)e = [Γ]T (f )e = [Γ]T [k ]e (δ ) e = [Γ]T [k ]e [Γ](∆ ) e [K11 ]e [K12 ]e  (F) = [K ] (∆ ) where [K ] =  e e [K 21 ] [K 22 ]  e

z

e

e

e

Nodal Equilibrium & Compatibility The same as the truss problems.

z

Global Stiffness Matrix (P) = [C]T (F) = [C]T [K ](∆) = [C]T [K ][C](u) (P) = [K ](u)

z

where

[K ] = [C]T [K ][C]

Direct Stiffness Method The same as the truss problems.


79


Chapter 6 Buckling of Structures

P

P

θ

L


80

School of Civil, Urban & Geosystem Eng., SNU 6.0 Stability of Structures

Q

Q

θ

L

z

Stable state

KθL × L > QθL → KL > Q The structure would return its original equilibrium position for a small perturbation in θ.

z

Critical state

KθL × L = QθL → KL = Q

z

Unstable state

KθL × L < QθL → KL < Q The structure would not return its original equilibrium position for a small perturbation in θ.


81

School of Civil, Urban & Geosystem Eng., SNU 6.1 Governing Equation for a Beam with Axial Force q

M

M+dM

w

Q V+dV

V

dw dx dx Q

z

Equilibrium for vertical force (V + dV ) − V + qdx = 0 →

z

Equilibrium for moment ( M + dM ) − M − Vdx + qdx

z

dV = −q dx

dx dw dM dw −P =0→ −Q =V dx dx dx 2

Elimination of shear force

d 2M d 2w − Q = −q dx 2 dx 2 z

Strain-displacement relation ε=−

z

d 2w du y+ 2 dx dx

Stress-strain relation (Hooke law) σ = Eε = − E

z

Definition of Moment

M = ∫ σydA = ∫ EεydA = − ∫ ( E A

z

d 2w du y+E 2 dx dx

A

A

d 2w 2 du d 2w − ) = − y E y dA EI dx 2 dx dx 2

Beam Equation with Axial Force

EI

d 4w d 2w + Q =q dx 4 dx 2


82

School of Civil, Urban & Geosystem Eng., SNU 6.2 Homogeneous Solutions z

Characteristic Equation for P > 0 w = e λx λx

e (λ + β λ ) = 0 → λ = ±β i, 0 z

4

2 2

where β 2 =

Q EI

Homogeneous solution

w = Aeβix + Be −βix + Cx + D z

Exponential Function with Complex Variable

i 2 2 i3 3 i 4 4 i5 5 i6 6 x + x + x + x + x +L 2! 3! 4! 5! 6! 2 3 4 (−i ) 2 (−i ) 3 (−i ) 4 (−i )5 5 (−i )6 6 e − ix = 1 − ix + x + x + x + x + x +L 2! 3! 4! 5! 6! 1 1 1 eix + e − ix = 2(1 − x 2 + x 4 − x 6 + L) = 2 cos x 2! 4! 6! i i i eix − e − ix = 2(ix − x 3 + x 5 − x 7 + L) = 2i sin x 3! 5! 7!

eix = 1 + ix +

eix = cos x + i sin x , e − ix = cos x − i sin x z

Homogeneous solution

w = A(cos βx + i sin βx) + B(cos β x − i sin βx) + Cx + D = ( A + B) cos β x + i ( A − B) sin β x + Cx + D = A cos β x + B sin β x + Cx + D z

Characteristic Equation for P < 0 w = e λx λx

e (λ − β λ ) = 0 → λ = ±β, 0 z

4

2 2

where β 2 =

Q EI

Homogeneous solution for P < 0

w = Aeβx + Be −βx + Cx + D eβ x + e − β x eβ x − e − β x = ( A + B) + ( A − B) + Cx + D 2 2 = A cosh β x + B sinh β x + Cx + D


83

School of Civil, Urban & Geosystem Eng., SNU Simple Beam

z

Q

Q

− Boundary Condition w(0) = A + D = 0 , w′′(0) = − Aβ2 = 0 → A = 0 w( L) = B sin β L + CL = 0 , w′′( L) = − Bβ2 sin β L = 0 → B = C = 0

− Characteristic Equation A = B = C = D = 0 → w = 0 (trivial solution) or n 2 π 2 EI βL = nπ → Q = , n = 1,2,3L L2 nπ w = B sin βx = B sin x L z

Fixed-Fixed Beam

Q

Q

− Boundary Condition w(0) = A + D = 0 w′(0) = β B + C = 0 w( L) = A cos βL + B sin βL + CL + D = 0 w′( L) = − Aβ sin βL + Bβ cos βL + C = 0 − Characteristic Equation 1 0   0 β   cos βL sin βL  − β sin β L β cos β L

1 0 0 1 0 1 A   0         0 1 0 1 0  B   0  β  )=0 → Det ( =  cos βL sin βL L 1 L 1 C   0        1 0 D   0  − β sin βL β cos βL 1 0

1 0 0 1 0 1 1 0 β β 0 1 0 β sin βL L = sin βL L 1 − cos βL cos βL sin βL L 1 β cos β L 1 0 − β sin βL β cos βL 1 − β sin βL β cos βL 1 0



84

− β − (−β cos β L) − (−β(cos βL + β L sin β L) + β) = β(2 cos β L − 2 + β L sin βL) = 0 2 cos β L − 2 + β L sin β L = 2(cos βL − 1) + βL sin βL = βL βL βL + 2β L sin cos = 2 2 2 βL βL βL βL βL βL βL βL sin ( cos − sin ) = 0 → sin = 0 or cos − sin =0 2 2 2 2 2 2 2 2 − 4 sin 2

− Eigenvalues Symmetric modes sin

βL βL 4n 2 π 2 EI =0 → = nπ → Q = , n = 1,2,3L 2 2 L2

w(0) = A + D = 0 , w′(0) = w′( L) = β B + C = 0, w( L) = A + CL + D = 0 A + D = 0 → A = − D → w = A(cos

2nπ x − 1) for A ≠ 0 L

Anti-symmetric modes βL βL βL βL βL 8.18π 2 EI cos − sin =0→ = tan →Q = 2 2 2 2 2 L2



85

Cantilever Beam Q

Q

− Boundary Condition w(0) = A + D = 0 w′(0) = β B + C = 0 M ( L) = − EIw′′( L) = − EI (− Aβ 2 cos β L − Bβ 2 sin β L) = 0 V ( L) = − EI Q=

d 3w dw −P =0 3 dx dx

n 2 π 2 EI , n = 1,2,3L 4 L2

6.3.Homogeneous and Particular solution w = wh + wp = A cos β x + B sin β x + Cx + D + wp EI

z

d 4 ( wh + w p ) dx 4

+Q

d 2 ( wh + w p ) dx 2

d 4 wp d 2wp d 4 wh d 2 wh = EI +Q + EI +Q dx 4 dx 2 dx 4 dx 2 d 4wp d 2 wp = EI +Q =q dx 4 dx 2

Four Boundary Conditions for Simple Beams w(0) = A + D + w p (0) = 0 , M (0) = − EIw′′(0) = − EI (− Aβ 2 + w′p′ (0)) = 0 w( L) = A cos β L + B sin βL + CL + D + w p ( L) = 0 M ( L) = − EIw′′( L) = − EI (− Aβ 2 cos β L − Bβ 2 sin β L + w′p′ ( L)) = 0 1 0 0 1   A   w p ( 0)    B   w′′ (0)   − β2 0 0 0   +  p  = 0 → KX + F = 0   cos βL L 1 C   wp ( L)  sin βL      2 2  − β cos β L − β sin β L 0 0 D   w′p′ ( L) 

z

The homogenous solution is for the boundary conditions, while the particular solution is for the equilibrium.


86

School of Civil, Urban & Geosystem Eng., SNU 6.4.Energy Method z

Total Potential Energy l

l

l

dw dw 1 d 2w d 2w 1 Π = ∫ 2 EI 2 dx − Q ∫ dx − ∫ wqdx 2 0 dx 2 0 dx dx dx 0

∆

L−∆

L=

∫ ds = ∫ 0

L−∆

∫

L= L−∆+

0

z

L−∆

1 + ( w′) dx ≈

L−∆

2

0

1 ( w′) 2 dx → ∆ = 2

∫ 0

L−∆

∫ 0

Q

1 (1 + ( w′) 2 )dx 2 L

1 1 ( w′) 2 dx ≈ ∫ ( w′) 2 dx for ∆ << L 2 20

Principle of the Minimum Potential Energy 1 d 2 (w e + w ) d 2 (we + w ) 1 d (we + w ) d (we + w ) EI dx Q dx − ∫ ( w e + w )qdx − 2 2 ∫ 2 ∫0 2 dx dx dx dx 0 0 l

Πh =

l

l

l

l

l

1 d 2 we d 2 we d 2 w e dw e dw e d 2 w dw e dw e )dx − ∫ w qdx + ∫ ( )dx EI Q EI Q = ∫( − − 2 0 dx 2 dx dx dx dx dx 2 dx 2 dx 2 0 0 l

− ∫ w qdx + 0

l

1 d 2w d 2 w dw dw ( )dx EI Q − 2 ∫0 dx 2 dx dx dx 2

l

l

d 4 we d 2 we 1 d 2w d 2 w dw dw Q q ) dx ( EI Q )dx = Π + ∫ w ( EI + − + − 2 ∫0 dx 2 dx dx dx 4 dx 2 dx 2 0 e

l

= Πe + z

1 d 2w d 2 w dw dw ( EI − Q )dx 2 ∫0 dx 2 dx dx dx 2

The principle of minimum potential energy holds if and only if l

∫( 0

z

d 2w d 2 w dw dw EI − Q )dx > 0 for all possible w dx dx dx 2 dx 2

The principle of the minimum potential energy is not valid for the following cases. l

∫( 0

d 2w d 2 w dw dw EI − Q )dx ≤ 0 for some w dx dx dx 2 dx 2


87


The critical status of a structure is defined as l

∫( 0

z

d 2w d 2 w dw dw EI − Q )dx = 0 dx dx dx 2 dx 2

Approximation n

− Approximation of displacement: w = ∑ ai g i i =1

− Critical Status l

n

n

l

n

0

i =1

n

l

n

∫ (∑ ai g i′′)EI (∑ a j g ′j′ )dx − P ∫ (∑ ai g i′ )(∑ a j g ′j )dx = i =1

0

n

n

j =1

l

n

j =1

n

n

n

n

∑∑ ai ∫ g i′′EIg i′′dxa j − Q∑∑ ai ∫ g i′g ′j dxa j = ∑∑ ai K ij a j − Q∑∑ ai K ijG a j = i =1 j =1

i =1 j =1

0

i =1 j =1

0

i =1 j =1

(a ) (K − K )(a ) = 0 → Det (K − K ) = 0 T

z

G

T

G

Example - Simple Beam Q

− with a parabola: w = ax( x − l ) → g1′ = 2 x − l , g1′′ = 2 l

l

0

0

l

l

∫ g i′′EIg i′′dx = ∫ 2 EI 2dx = 4 EIl l

4 1 3 2 2 2 3 ∫0 g i′EIg i′dx = ∫0 (2 x − l ) dx = ∫0 (4 x − 4 xl + l )dx = ( 3 − 2 + 1)l = 3 l 1 12 EI Det (4 EIl − Q l 3 ) = 0 → Qcr = 2 3 l

− with one sine curve: w = a sin

(exact :

EI π 2 EI = 9.86 2 , error = 22%) 2 l l

πx πx π πx π → g1′ = cos , g1′′ = ( ) 2 sin l l l l l

π 4 π 4 l 2 πx ∫0 g i′′EIg i′′dx = EI ( l ) ∫0 sin l dx = EI ( l ) 2 l

l

π 2 π 2 l 2 πx ∫0 g i′EIg i′dx = ( l ) ∫0 cos l dx = ( l ) 2 l

l

π l π l π 2 EI Det ( EI ( ) 4 − Q( ) 2 ) = 0 → Q = 2 (exact ) l 2 l 2 l


88


Example – Cantilever Beam Q

Q

− with one unknown: w = ax 2 → g1′ = 2 x , g1′′ = 2 l

l

0

0

∫ g i′′EIg i′′dx = ∫ 2 EI 2dx = 4EIl ,

l

l

0

0

2 ∫ g i′EIg i′dx = ∫ 4 x dx =

4 3EI Det (4 EIl − Q l 3 ) = 0 → Qcr = 2 3 l

4 3 l 3

π 2 EI EI (exact : = 2.46 2 , error = 22%) 2 4l l

− with two unknowns: w = ax 2 + bx 3 → g1′ = 2 x , g 2′ = 3 x 2 , g1′′ = 2 , g 2′′ = 6 x l

l

l

4 6 9 G G = ∫ 6 x 3dx = l 4 , K 22 = ∫ 9 x 4 dx = l 5 K11G = ∫ 4 x 2 dx = l 3 , K12G = K 21 3 4 5 0 0 0 l

l

l

K11 = ∫ 4dx = 4l , K12 = K 21 = ∫ 12 xdx = 6l , K 22 = ∫ 36 x 2 dx = 12l 3 2

0

0

 4l 6l 2  1  40l 3 Det ( EI  2 −Q  4 3 30 45l 6l 12l 

0

45l 4  4l − 40l 3 α 6l 2 − 45l 4 α 1 Q = 0,α = )=0→ 2 5 4 3 5 30 EI 6l − 45l α 12l − 54l α 54l 

(4l − 40l 3 α)(12l 3 − 54l 5 α) − (6l 2 − 45l 4 α) 2 = 0 → 4l 4 − 52l 6 α + 45l 8 α 2 = 0 EI EI 26l 6 ± 26 2 l 12 − 180l 12 26 ± 22.27 0.0829 1.0727 α= = = or → Qcr = 2.487 2 or 32.181 2 8 2 2 2 l l l l 45l 45l

Qexact = 2.49

EI EI (error = 1.2%) or Qexact = 22.19 2 (error = 45%) 2 l l


89

School of Civil, Urban & Geosystem Eng., SNU 6.5 Approximation with the Homogeneous Beam Solutions z

Homogeneous Solution of Beam u2i −1

u 2 i +1 P

u2 i + 2

u2i i-th member

 δ Ly     θ Lz  i L L R R w = N 1δ y + N 2 θ z + N 3 δ y + N 4 θ z = ( N 1 , N 2 , N 3 , N 4 ) R  = ( N )(∆ i ) = (N)(∆ i ) δy   R  θz  2 3 2 3 2 x 2 x3 3x 2x 2x x 3x 2 x3 N1 = 1 − 2 + 3 , N 2 = x − + 2 , N3 = − 2 , N4 = − + 2 L L L L L L L L 6.5.1 Beam Analysis z

Total Potential Energy l l p p l d 2 wi dwi dwi 1 d 2 wi 1 Π=∑ ∫ EI dx − ∑ Q ∫ dx − ∑ ∫ wi qi dx 2 dx dx dx 2 i =1 2 i =1 0 i =1 2 0 dx 0 p

p

=∑ i =1

l l p p l d 2 wi dwi T dw i 1 d 2 wi T 1 − − ( ) ( ) ( wi ) T qi dx EI dx Q dx ∑ ∑ 2 ∫ ∫ 2 ∫0 dx 2 2 dx dx dx i =1 i =1 0 0 l

l

l

p p 1 1 d 2N T d 2N dN T dN T T = ∑ (∆ i ) ∫ ( 2 ) EI dx(∆ i ) − ∑ Q(∆ i ) ∫ ( ) dx(∆ i ) − ∑ (∆ i ) T ∫ (N) T qi dx 2 dx dx dx i =1 2 i =1 2 i =1 0 dx 0 0 p

=

p 1 p 0 T G ∆ − ∆ − ( ) ([ ] [ )( ) (∆ i ) T (f i ) K K ∑ i ∑ i i i 2 i =1 i =1

=

p 1 T p (u) ∑ [C i ]T ([K i0 ] − [K Gi ])[C i ](u) − ∑ (u) T [C i ]T (f i ) 2 i =1 i =1

p 1 T p (u) ∑ [C i ]T [K i ][C i ](u) − (u) T ∑ [C i ]T (f i ) 2 i =1 i =1 1 T = (u) [K ](u) − (u) T (P ) 2

=

 12  L2  i  6 EI  Li K i0 = i  Li − 12 2  Li  6   Li

6 Li 4 −

6 Li 2

12 L2i 6 − Li 12 L2i 6 − Li

−

6 Li   2  , 6 −  Li   4 

 6  5L i  1  K Gi = Q  10 6 −  5 Li 1   10

1 10 2 Li 15 1 10 L − i 30

6 5Li 1 10 6 5Li 1 − 10

−

1 10   Li  − 30  1 −  10  2 Li  15 


90


Principle of Minimum Potential Energy for Q < Qcr Π=

n n 1 T 1 n (u) [K ](u) − (u)T (P) = ∑ ui ∑ K ij u j − ∑ ui Pi 2 2 i=1 j =1 i =1

∂Π 1 n ∂u i = ∑ ∂u k 2 i =1 ∂u k

z

n

∑ K ij u j + j =1

n n ∂u j 1 n u K − u i Pi ∑ i ∑ ij ∂u ∑ 2 i =1 j =1 i =1 k

n

=

1 1 n 1 n 1 n K u + u K − P = K u + K ki u i − Pk ∑ kj j 2 ∑ ∑ kj j 2 ∑ i ik k 2 j =1 2 j =1 i =1 i =1

=

n 1 n 1 n K kj u j + ∑ K kj u j − Pk = ∑ K kj u j − Pk = 0 for k = 1, L , n → [K ](u) = (P) ∑ 2 j =1 2 j =1 j =1

Calculation of the Critical Load

Det ([K ]) = Det ([K 0 ] − [K G ]) = 0 z

Frame Members

 f xL       L  fy       mL    E  =(  f R  Le  x     R  fy       mR   

 Ae  0  0    Ae 0   0 

0 12 I e L2e 6Ie Le 0 12 I e − 2 Le 6Ie Le

0 6Ie Le

Ae

4Ie

0

0 6Ie − Le

Ae

2Ie

0

0

0

0 12 I e − 2 Le 6Ie − Le 0 12 I e L2e 6I − e Le

 0 0  6   0  5 Le   1 2 I e  0 10  − P  0 0 0  6  6Ie  0 − −  5Le Le    1 0 4Ie   10  0 6Ie Le

P = EA

0 1 10 2 Le 15 0 1 10 L − e 30

0 0 0 0 0 0

0 6 − 5 Le 1 10 0 6 5Le 1 − 10

 δ Lx      0   δL  1   y   10     L L − e   θe  30  ) 0   1   δR  −  x 10    2 Le   R  δ 15   y       θR   e

δ Lx − δ Rx Le


91

School of Civil, Urban & Geosystem Eng., SNU 6.6 Nonlinear Analysis of Truss R f yR δ y

f yL δ Ly

L f xL δ x

z

f xR δ Rx

Force – Displacement relation at Member ends EA R (δ x − δ Lx ) L EA R f xR = (δ x − δ Lx ) L δ Ry − δ Ly R L R fy = − fy = fx l f xL = −

z

Member Stiffness Matrix       

e

 1 f xL    L fy  EA   0 = f xR  l  − 1  f yR    0

0 − 1 0 0 0 0 δ Rx − δ Lx + 0 1 0 l  0 0 0

0 0 0 1  0 0  0 − 1

0  δ Lx    0 − 1  δ Ly    0 0  δ Rx   0 1  δ Ry  0

e

(f )e = ([k ]e0 + f xR [k ]eg )(δ )e = ([k ]e0 + p e [k ]eg )(δ )e z

Equilibrium Analysis (f )ie = ([k ]e0 + pie [k ]eg )(δ )ie (F)e = [Γ]T (f )e = [Γ]T [k ]e (δ )e = [Γ]T ([k ]0e + pie [k ]eg )[Γ](∆ )e = [K ]e (∆ )e

z

Successive substitution (F) ie−1 ≈ [Γ]T ([k ]e0 + pie−1 [k ]eg )[Γ](∆ ) e = [K ]ie−1 (∆ ) e

(P) = ([K ]0 + [K ( pie−1 )]G )(u)i


92


Newton-Raphson Method

(f )ie = ([k ]e0 + pie [k ]eg )(δ )ie = ([k ]e0 + ( pie−1 + ∆pie )[k ]eg )(δ ie−1 + ∆δ ie ) = ([k ]0e + pie−1[k ]eg )δ ie−1 + ([k ]0e + pie−1[k ]eg )∆δ ie + ∆pie [k ]eg (δ ie−1 + ∆δ ie ) ≈ ([k ]e0 + pie−1[k ]eg )δ ie−1 + ([k ]e0 + pie−1[k ]eg )∆δ ie + ∆pie [k ]eg δ ie−1 = (f )ie−1 + ([k ]e0 + pie−1[k ]eg + [k ]eσ )∆δ ie ([k ]0e + pie−1[k ]eg + [k ]eσ )∆δ ie = (f )ie − (f )ie−1 = (∆f )ie

([K ]0 + [K ( pie−1 )]G + [K ]σ )∆u = (∆P)i

∆pie [k ]eg δ ie−1

0 0 0 1 EA  = 0 l 0  0 − 1   L EA  δ y = 2  l  δR  y

e

0 δ Lx    0 − 1 δ Ly  ∆δ Rx − ∆δ Lx   0 0 δ Rx  l  R  0 1 δ y  i −1 0

e

e

 ∆δ Lx    L  − δ Ry   ∆δ y  ( ) 1 0 1 0 −  R 0   ∆δ x   L   ∆δ R  − δy   y i i −1 0

 0  EA R L e  1 = 2 (δ y − δ y ) i −1  0 l  − 1

e

0 0 ∆δ Lx    0 − 1 0 ∆δ Ly    0 0 0 ∆δ Rx   0 1 0 ∆δ Ry  i 0

= [k ]eσ ∆δ ie


Class Notes on Structural Analysis-2 for Stiffness & Flexibility Method of Matrix By HSL

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