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 Compar Compariso ison n of Flexibi Flexibility lity Method Method and and Stiffne Stiffness ss Metho Method…… d…………… ……………… …………… …… 2 1.1 Analysi Analysiss of Fundam Fundament ental al System System……… ……………… ……………… ……………… ……………… ……………… ………... ..... .. 5 1.2 1.2 Ana Analy lysis sis of Beam Beams… s……… ………… ………… ………… ………… ………… ………… ………… ………… ………… ………… ………… …… 8 1.3 Analy Analysis sis of Frame Frames…… s…………… ……………… ……………… ……………… ……………… ……………… ……………… ………….. ….... 17 Chapter 2 Iterative Solution Method & Moment Distribution Method
2.1 Solut Solution ion Method Method for for Linear Algebr Algebraic aic Equatio Equations… ns………… ……………… ……………… ……………… ……… 2.2 Moment Moment Distribution Distribution Method……………… Method……………………………… ……………………………… ……………………... ……... 2.3 Example Example - MDM for for a 4-span Continuous Continuous Beam………………… Beam………………………………… ……………….. 2.4 Direct Solution Solution Scheme Scheme by by Partitioning Partitioning……………… ……………………………… ………………………….. ………….... 2.5 Momen Momentt Distribu Distributio tion n Method Method for Frame Frames…… s…………… ……………… ……………… ……………… …………… …… Chapter 3 Energy Principles
32
33 37 42 44 45 47
3.1 Spring-F Spring-Forc orcee System Systems……… s……………… ……………… ……………… ……………… ……………… ……………… ……………. ……... .. 48 3.2 Beam Beam Problem Problems…… s…………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ……….. 49 3.3 Truss Truss prob problem lems……… s……………… ……………… ……………… ……………… ……………… ……………… ……………… ……………. ……. 53 Chapter 4 Matrix Structural Analysis
4.1 Truss Truss Proble Problems… ms………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ………… … 4.2 Beam Beam Problem Problems…… s…………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ……… 4.3 Frame Frame Proble Problems… ms………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ………... ... Chapter 5 Buckling of Structures
5.0 Stabil Stability ity of Structur Structures…… es…………… ……………… ……………… ……………… ……………… ……………… ……………… ………... ... 5.1 Governi Governing ng Equatio Equation n for a Beam with Axial Force…………………… Force………………………………. …………... .. 5.2 Homogen Homogeneou eouss Soluti Solutions ons……… ……………… ……………… ……………… ……………… ……………… ……………… ………….. ….... 5.3 Homogeneo Homogeneous us and Particular Particular solution……… solution……………………… ……………………………… …………………….. …….. 5.4 Ener Energy gy Metho Method…… d…………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ……….. 5.5 Approximati Approximation on with the Homogeneou Homogeneouss Beam Beam Solutio Solutions………… ns………………………… ……………….. 5.6 Nonli Nonlinea nearr Analysis Analysis of Truss…… Truss…………… ……………… ……………… ……………… ……………… ……………… …………. ….
56
57 68 76 79
80 81 82 85 86 89 91
Contents
Chapter 1 Slope Deflection Method
1
1.0 Compar Compariso ison n of Flexibi Flexibility lity Method Method and and Stiffne Stiffness ss Metho Method…… d…………… ……………… …………… …… 2 1.1 Analysi Analysiss of Fundam Fundament ental al System System……… ……………… ……………… ……………… ……………… ……………… ………... ..... .. 5 1.2 1.2 Ana Analy lysis sis of Beam Beams… s……… ………… ………… ………… ………… ………… ………… ………… ………… ………… ………… ………… …… 8 1.3 Analy Analysis sis of Frame Frames…… s…………… ……………… ……………… ……………… ……………… ……………… ……………… ………….. ….... 17 Chapter 2 Iterative Solution Method & Moment Distribution Method
2.1 Solut Solution ion Method Method for for Linear Algebr Algebraic aic Equatio Equations… ns………… ……………… ……………… ……………… ……… 2.2 Moment Moment Distribution Distribution Method……………… Method……………………………… ……………………………… ……………………... ……... 2.3 Example Example - MDM for for a 4-span Continuous Continuous Beam………………… Beam………………………………… ……………….. 2.4 Direct Solution Solution Scheme Scheme by by Partitioning Partitioning……………… ……………………………… ………………………….. ………….... 2.5 Momen Momentt Distribu Distributio tion n Method Method for Frame Frames…… s…………… ……………… ……………… ……………… …………… …… Chapter 3 Energy Principles
32
33 37 42 44 45 47
3.1 Spring-F Spring-Forc orcee System Systems……… s……………… ……………… ……………… ……………… ……………… ……………… ……………. ……... .. 48 3.2 Beam Beam Problem Problems…… s…………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ……….. 49 3.3 Truss Truss prob problem lems……… s……………… ……………… ……………… ……………… ……………… ……………… ……………… ……………. ……. 53 Chapter 4 Matrix Structural Analysis
4.1 Truss Truss Proble Problems… ms………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ………… … 4.2 Beam Beam Problem Problems…… s…………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ……… 4.3 Frame Frame Proble Problems… ms………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ………... ... Chapter 5 Buckling of Structures
5.0 Stabil Stability ity of Structur Structures…… es…………… ……………… ……………… ……………… ……………… ……………… ……………… ………... ... 5.1 Governi Governing ng Equatio Equation n for a Beam with Axial Force…………………… Force………………………………. …………... .. 5.2 Homogen Homogeneou eouss Soluti Solutions ons……… ……………… ……………… ……………… ……………… ……………… ……………… ………….. ….... 5.3 Homogeneo Homogeneous us and Particular Particular solution……… solution……………………… ……………………………… …………………….. …….. 5.4 Ener Energy gy Metho Method…… d…………… ……………… ……………… ……………… ……………… ……………… ……………… ……………… ……….. 5.5 Approximati Approximation on with the Homogeneou Homogeneouss Beam Beam Solutio Solutions………… ns………………………… ……………….. 5.6 Nonli Nonlinea nearr Analysis Analysis of Truss…… 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
k 1
P
k 2
k 1
Remove redundancy (Equilibrium)
k 2
Compatibility
P P
X δ1 = δ 2 = δ
Compatibility
Equilibrium
δ1 = δ 2
X k 1
=
P − X k 2
→ X =
k 1δ + k 2δ = P → δ = k 1 k 1 + k 2
P
P k 1 + k 2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
3
School of Civil, Urban & Geosystem Eng., SNU Flexibility Method
B
A
P
L/2
P
L/2
Stiffness Method
EI
EI
L
L
Remove redundancy (Equilibrium)
B
A
C
EI
EI
L
L
C
Compatibility
3 PL 16
+
θ B
+
1
θ BA = θ BC = θ B
Compatibility δ B 0
L
2
1 PL PL 2 L = ×1 = (1 + ) , δ BB = 6 EI 2 4 16 EI 3 EI
δ B 0 + δ BB M B = 0 → M B = −
δ B 0 δ BB
=−
3 PL 32
Equilibrium f =− M BA
3 PL
f B B = 0 , M BA = M BC = , M BC
3 EI
16 L f f B B = + + + = M M M M M 0 ∑ B BA BC BA BC → −
3 PL 16
+
6 EI L
θ B = 0 → θ B =
θ B
PL2 32 EI
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
3
School of Civil, Urban & Geosystem Eng., SNU Flexibility Method
B
A
P
L/2
P
L/2
Stiffness Method
EI
EI
L
L
C
Remove redundancy (Equilibrium)
B
A
EI
EI
L
L
C
Compatibility
3 PL 16
+
θ B
+
1
θ BA = θ BC = θ B
Compatibility δ B 0
f =− M BA
2
L
δ B 0 δ BB
=−
3 PL
f B B = 0 , M BA = M BC = , M BC
3 EI
16 L f f B B ∑ M B = M BA + M BC + M BA + M BC = 0 →
1 PL PL 2 L = ×1 = (1 + ) , δ BB = 6 EI 2 4 16 EI 3 EI
δ B 0 + δ BB M B = 0 → M B = −
Equilibrium
3 PL
−
32
3 PL 16
+
6 EI L
θ B = 0 → θ B =
θ B
PL2 32 EI
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
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 stiffness equation.
6.
Calculate reactions and other quantities as needed.
5. Solve the flexibility equation. 6. Calculate reactions and other quantities as needed.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
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 stiffness equation.
6.
Calculate reactions and other quantities as needed.
5. Solve the flexibility equation. 6. Calculate reactions and other quantities as needed.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
5
School of Civil, Urban & Geosystem Eng., SNU 2.1 Analysis of Fundamental System 2.1.1 En d Rotation
A
B θ A
Flexibility Method i) θ B = 0 M A
M B
L
L M A + M B = −θ A 3 EI 6 EI L L M A + M B = 0 6 EI 3 EI ii)
θ A =
→ M A = − 4 EI θ A , L
0 2 EI
4 EI
M B =
2 EI L
θ A
5
School of Civil, Urban & Geosystem Eng., SNU 2.1 Analysis of Fundamental System 2.1.1 En d Rotation
A
B θ A
Flexibility Method i) θ B = 0 M A
M B
L
L M A + M B = −θ A 3 EI 6 EI L L M A + M B = 0 6 EI 3 EI ii)
θ A =
→ M A = − 4 EI θ A , L
M B =
2 EI L
θ A
0 M A = −
2 EI L
θ B , M B =
Sign Convention for M :Counterclockwise “+”
θ A ≠ 0 , θ B ≠ 0
M A = M B =
4 EI L 2 EI L
θ A + θ A +
4 EI
2 EI L 4 EI L
L
θ B
θ B θ B
2.1.2 Rel ative moti on of j oin ts
∆
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
6
School of Civil, Urban & Geosystem Eng., SNU
Flexibility Method
∆
L ∆ M A + M B = − 3 EI 6 EI L ∆ L L M A + M B = 6 EI 3 EI L L
→ M A = − 6 EI ∆ , L L
or in the new sign convention : M A =
L L
, M B =
6 EI ∆ L L 6 EI ∆ L L
Final Slope-Deflection Equation
M A = M B =
6 EI ∆
M B =
4 EI L 2 EI L
θ A + θ A +
2 EI L 4 EI L
θ B + θ B +
6 EI ∆ L L 6 EI ∆ L L
In Case an One End is Hinged
M A = M B =
4 EI L 2 EI L
θ A + θ A +
2 EI L 4 EI L
θ B + θ B +
6 EI ∆ L L 6 EI ∆ L L
=0→ =
3 EI L
2 EI L θ B +
EI
θ A = −
L
θ B −
3 EI ∆ L L
3 EI ∆ L L
2.1.3 F ix ed End F orce
Both Ends Fixed
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
7
School of Civil, Urban & Geosystem Eng., SNU
One End Hinged
M B
M A +
M A
M A/2 3 M A 2
3 M A
Ex.: Uniform load case with a hinged left end f B
M = −
qL2 12
−
qL2 24
=−
3qL2 24
=−
qL2 8
M A f = 0
,
2.1.4 Joint Equil ibri um
F fixed F joint
F member
Joint i
− ∑ F fixed −∑ F member + ∑ F joint = 0
or
∑ F fixed +∑ F member = ∑ F joint Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
8
School of Civil, Urban & Geosystem Eng., SNU 2.2 Analysis of Beams 2.2.1 A F ix ed-fi xed En d Beam
P B
A
C
EI a
DOF : θ B ,
Analysis
b
B
i) All fixed : No fixed end forces ii) θ B ≠ 0 , ∆ B = 0
1 M AB =
2 EI a
1 1 V BA = −V AB =
2 EI a
a
6 EI a
2
2
a 1
4 EI a
6 EI
θ B
a
2
1
θ B , M BC =
6 EI
1
θ B , − V BC = V CB =
θ B
6 EI
4 EI
1
θ B , M BA =
b2
4 EI b
2 EI b
θ B
θ B
4 EI
θ B
1
θ B , M CB =
b
2 EI
θ B
b
6 EI
θ B
b
2
θ B
6 EI θ B
b2
θ B
iii) θ B = 0 , ∆ B ≠ 0
2 M AB =
6 EI a2
2 2 V BA = −V AB =
6 EI a
2
2
∆ B , M BA =
12 EI a3
a
3
2
6 EI a
∆ B
12 EI a
a2
2
∆ B , M BC = − 2
∆ B , V BC = −V CB =
∆ B
12 EI
6 EI
3
2
∆ B
∆ B
12 EI b2 b
b2
2
∆ B M CB = −
6 EI b2
∆ B
∆ B
6 EI 2
6 EI
6 EI
∆ B
b2
12 EI b3
∆ B
∆ B
12 EI b3
∆ B
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
9
School of Civil, Urban & Geosystem Eng., SNU
Construct the Stiffness Equation
∑ M B i
1
1
2
2
= 0 → M BA + M BC + M BA + M BC = 0 → 4 EI (
i 1 1 2 2 ∑ V B = P → V BA + V BC + V BA + V BC = P →
6 EI (
1 a 1
+
1 b
)θ B + 6 EI (
1 a
2
1
−
)θ B + 12 EI ( 2
a2 b (b − a )a 2b 2 a 3b3 P , ∆ B = P θ B = − 2 EIl 3 3 EIl 3 1 AB
M AB = M
2 AB
+ M
=
1 2 M CB = M CB + M CB =
2 EI a 2 EI b
θ B +
θ B −
6 EI a2 6 EI b2
∆ B =
1
−
b2
1 a3
+
)∆ B = 0 1 b3
)∆ B = P
ab 2
P , l 2 a 2b ∆ B = − 2 P l
2.2.2 Anal ysis of a Two-span Conti nu ous Beam (Approach I )
q
qL B
A
C
EI
2 EI
L
DOF : θ B ,
Analysis
L
C
i) Fix all DOFs and Calculate FEM. f AB
=
M
qL2 12
f BA
, M
=−
qL2 12
f BC
, M
=
qL2 8
f CB
, M
qL2
=−
8
ii) θ B ≠ 0 , θC = 0 2 EI
1 = M AB
L
4 EI
1
θ B , M BA =
L
1
θ B , M BC =
8 EI L
1
θ B , M CB =
4 EI L
θ B
iii) θ B = 0 , θC ≠ 0 2 = M BC
4 EI L
2
θC , M CB =
8 EI L
θC
Construct the Stiffness Equation
∑ M
i B
f BA
= 0 → M
∑ M = 0 → M i C
f CB
f BC
+ M
1 CB
+ M
1 BA
+ M
2 CB
+ M
1 BC
+ M
=0→
2 BC
+ M
=0→
qL2 24 −
EI
+ 12
qL2 8 L
L
EI
+4
L
EI
θ B + 4
L
EI
θ B + 8
L
θ C = 0
θC = 0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
10
School of Civil, Urban & Geosystem Eng., SNU θ B = −
96 EI
, θC =
qL3 48 EI
Member End Forces f AB
1 AB
M AB = M
+ M
f BA
=
1 BA
qL2
f BC
M BC = M
1 BC
+ M
f CB
+
12
M BA = M + M = −
qL2 12 2 BC
+ M
1 CB
2 CB
M CB = M + M + M
qL3
2 EI L
+
=
θ B =
4 EI L qL2 8
=−
48
qL2 1
2
θ B = − qL
8
+
qL2 8
3
8 EI L
+
θ B +
4 EI L
4 EI L
θ B +
θC =
8 EI L
1 8
qL2
θC = 0
Various Diagram
- Freebody Diagram
1 16
1
qL2
8
7 16
9 qL
16
qL2
5
qL
8
19 16
qL
3 8
qL
qL
- Moment Diagram
1 16
qL2
17 512
qL2
1 8
qL2
3 16
qL2
7
L 16
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
11
School of Civil, Urban & Geosystem Eng., SNU 2.2.3 Anal ysis of a Two-span Conti nu ous Beam (Appr oach I I )
q
qL B
A
C 2 EI
EI L
DOF : θ B
Analysis
L
i) Fix all DOFs and Calculate FEM. f AB
M
qL2
=
12
f BA
, M
=−
qL2 12
f BC
, M
=
qL2 8
+
1 qL2 2 8
=
3qL2 16
ii) θ B ≠ 0 1 M AB =
2 EI L
1
θ B , M BA =
4 EI L
1
θ B , M BC =
6 EI L
θ B
Construct Stiffness Equation
∑ M B qL2
f
f
1
1
= 0 → M BA + M BC + M BA + M BC = 0
EI EI qL2 + +4 θ B + 6 θ B = 0 → θ B = − 12 16 L L 96 EI
3qL2
Member End Forces f AB
M AB = M
1 AB
+ M
=
qL2
12 qL2
f 1 M BA = M BA + M BA = −
f 1 + M BC = M BC = M BC
+
+
12 3qL2 16
2 EI
θ B =
L 4 EI
+
48
qL2 1
2
θ B = − qL
L 6 EI L
3
θ B =
8 1 8
qL2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
12
School of Civil, Urban & Geosystem Eng., SNU ) 2.2.4 Anal ysis of a Beam with an I nternal H in ge (4 DOF s System
q C
B
A EI
EI
l
EI l
l
DOF : θ B , θ LC , θ RC , ∆C
Analysis
D
i) All fixed f AB
M
=
ql 2 12
f BA
, M
=−
ql 2 12
ii) θ B ≠ 0
1 M AB =
2 EI l
1
1
θ B , M BA = M BC =
4 EI l
1
θ B , M CB =
2 EI l
1
θ B , V CB =
6 EI l 2
θ B
iii) θ LC ≠ 0
2 M BC =
2 EI l
L
2
θC , M CB =
4 EI l
L
2
θC , V CB =
6 EI 2
l
L
θC
iv) θ RC ≠ 0
3 M CD =
4 EI l
R
3
θC , M DC =
2 EI l
R
3
θC , V CD = −
6 EI 2
l
L
θC
v) ∆ C ≠ 0
4 4 = M CB = M BC
6 EI l 2
4
4
∆ C , M CD = M DC = −
6 EI l 2
4
4
∆ C , V CD = V CB =
12 EI l 2
∆ C
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
13
School of Civil, Urban & Geosystem Eng., SNU
Construct Stiffness Equation
∑ M
i 1
=0→−
i
ql 2
EI
+8
i
∑ M 3 = 0 → i
i
=0→
i
θC +
l l EI EI L 2 θ B + 4 θC + l l
i
i
L
EI
∆ C = 0 l 2 EI 0 + 6 2 ∆ C = 0 l EI R EI 4 θC − 6 2 ∆ C = 0 l l EI EI EI EI 6 2 θ B + 6 2 θ LC − 6 2 θ RC + 24 3 ∆ C = 0 l l l l
12
∑ M 2 = 0 →
∑V 4
EI
θ B + 2
0
+6
Elimination of θ LC and θ RC
- 2nd and 3rd equation EI R EI EI L EI EI θC = −( θ B + 3 2 ∆ C ) , 2 θC = 3 2 ∆ C 2 l l l l l
- 1st equation −
ql 2
EI
+8
12
l
EI
θ B + 2
l
EI
L
θ C + 6
l 2
∆ C =
ql 2
EI EI EI EI ql 2 EI EI − +8 θ B − ( θ B + 3 2 ∆ C ) + 6 2 ∆ C = 0 → − +7 θ B + 3 2 ∆ C = 0 12 l l 12 l l l l th - 4 equation EI EI EI EI 6 2 θ B + 6 2 θ LC − 6 2 θ RC + 24 3 ∆ C = 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
2.2.5 An alysis of a Beam wit h an I ntern al H in ge (2 DOF s System) q
C
B
A EI
EI
l
DOF : θ B ,
Analysis
l
EI
D
l
C
i) All fixed f AB
M
=
ql 2 12
f BA
, M
=−
ql 2 12
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
14
School of Civil, Urban & Geosystem Eng., SNU ii) θ B ≠ 0
1 M AB =
2 EI l
1
θ B , M BA =
4 EI l
1
θ B , M BC =
3 EI
1
1
θ B , V BC = −V CB = −
l
3 EI l 2
θ B
iii) ∆ C ≠ 0
2 2 M BC = − M DC =
3 EI 2
l
2
2
∆ C , V BC = −V CB = −
3 EI 2
l
2
2
∆ C , V CD = −V DC =
3 EI l 2
∆ C
Construct the Stiffness Equation
∑ M
i 1
=0→−
i
∑V 2
i
ql 2 12
EI
+7
EI
∆ C = 0 l l 2 EI EI 3 2 θ B + 6 3 ∆ C = 0 l l
=0→
i
θ B + 3
θ B =
2 EI l
R θC
L
θC = −
=−
1 2
ql 3 66 EI
3 EI
, ∆ C = −
∆ C
R θC
(− )→ l l 3∆ ql 3
θ B −
2 L
=−
=
+
ql 4 132 EI
3 ∆ C 2 l 3 ql 3
=−
132 EI 2 132 EI
=
3 ql 3 264 EI ql 3 264 EI
2.2.6 Beam wi th a Spri ng Support
q A EI
EI
l
C
B
l
k
EI
D
l
Analysis
i) All fixed f AB
M
=
ql 2 12
f BA
, M
=−
ql 2 12
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
15
School of Civil, Urban & Geosystem Eng., SNU ii) θ B ≠ 0
1 M AB =
2 EI l
1
θ B , M BA =
4 EI l
1
θ B , M BC =
3 EI l
1
1
θ B , − V BC = V CB =
3 EI l 2
θ B
iii) ∆ C ≠ 0
k ∆C
2 2 M BC = − M DC = 2 2 = −V CB = − V BC
l 2 3 EI 2
l
∆ C 2
2
∆ C , V CD = −V DC =
∑ M = 0 → − i
∑V 2
i
ql 2 12
=0→
i
θ B =
1+ α
EI
+7
2
l
2
∆ C , V S = k ∆ C
EI
θ B
∆ C = 0 l l 2 EI EI 3 2 θ B + (6 3 + k ) ∆ C = 0 l l
ql 3
1 + 14α / 11 66 EI
+3
, ∆ C = −
ql 4
1
(1 + 14α / 11) 132 EI
where k = α
6 EI l 3
α→0 θ B =
3 EI
Construct the Stiffness Equation i 1
3 EI
ql 3 66 EI
, ∆ C = −
ql 4 132 EI
α→∞ θ B =
ql 3 84 EI
, ∆ C = 0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
16
School of Civil, Urban & Geosystem Eng., SNU 2.2.7 Suppor t Settl ement
l
l
B
A EI
C
EI δ
DOF : θ B
Analysis
i) All fixed f M BA =
6 EI δ l l
f , M BC =−
3 EI δ l l
ii) θ B ≠ 0 1 = M BA
4 EI l
3 EI
1
θ B , M BC =
l
θ B
Construct the Equilibrium Equation
EI δ
∑ M 1 = 0 → 6 i
i
l l
EI δ
−3
l l
EI
+4
l
EI
θ B + 3
l
θ B = 0 → θ B = −
3δ 7 l
2.2.8 Temper atur e Change
A
B T 1 T 2
θ A =
α(T 2 − T 1 )
2h
l ,
θ B = −
α(T 2 − T 1 )
2h
l
Fixed End Moment
M A =
4 EI
θ A +
2 EI
θ B =
α(T 2 − T 1 )
EI h α(T 2 − T 1 ) M B = EI θ A + θ B = − L L h L 2 EI
L 4 EI
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
17
School of Civil, Urban & Geosystem Eng., SNU 2.3 Analysis of Frames 2.3.1 A Portal F r ame with out Sidesway
P
l /2 B
C EI 2
EI 1
EI 1
A
DOF : θ B , θC
Analysis
i)
D
All fixed Pl Pl 0 0 M BC , M CB = =− 8 8
ii) θ B ≠ 0 1 M AB =
1 M BC =
2 EI 1 l 4 EI 2 l
1
θ B , M BA =
4 EI 1
1
θ B , M CB =
l
θ B
2 EI 2 l
θ B
iii) θC ≠ 0 2 M BC =
2 M CD =
2 EI 2 l 4 EI 1 l
2
θC , M CB = 2
θC , M DC =
4 EI 2 l 2 EI 1 l
θC θC
Construct the Stiffness Equation
∑ M B = 0 →
Pl
∑ M C = 0 → −
Pl
i
i
− θ B = θC =
8 8 1
+(
+
4 EI 1 l
2 EI 2
+
4 EI 2 l
θ B + (
l Pl 2
)θ B +
4 EI 1 l
+
2 EI 2 l 4 EI 2 l
θC = 0
)θC = 0
4 EI 1 + 2 EI 2 8
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
18
School of Civil, Urban & Geosystem Eng., SNU
Member End Forces
M AB = M BA = M BC =
2 EI 1 l 4 EI 1 l Pl 8
+
θ B = −
θ B = −
4 EI 2
2 EI 1
4 EI 1 + 2 EI 2 8 4 EI 1
Pl
4 EI 1 + 2 EI 2 8
θ B +
l
Pl
2 EI 2 l
θC =
4 EI 1
Pl
4 EI 1 + 2 EI 2 8
Pl 2 EI 2 4 EI 2 4 EI 1 Pl M CB = − + θ B + θC = − 8 l l 4 EI 1 + 2 EI 2 8 M CD = M DC =
4 EI 1 l 2 EI 1 l
4 EI 1
θC =
Pl
4 EI 1 + 2 EI 2 8 2 EI 1
θC =
Pl
4 EI 1 + 2 EI 2 8
In case EI 1 = EI 2
Pl Pl Pl Pl Pl M AB = − , M BA = − , M BC = , M CB = − , M CD = 24 12 12 12 12
, M DC =
Pl 24
2.3.2 A Por tal F r ame wi thout Sidesway –hi nged suppoor ts
P
l /2 B
C EI 2
EI 1
A
DOF : θ B , θC
Analysis
EI 1
D
i) All fixed 0 M BC =
Pl 8
0 , M CB =−
Pl 8
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
School of Civil, Urban & Geosystem Eng., SNU
19
ii) θ B ≠ 0 1 M BA = 1 M BC =
3 EI 1 l 4 EI 2 l
θ B
2 EI 2
1
θ B , M CB =
l
θ B
iii) θC ≠ 0 2 = M BC 2 M CD =
2 EI 2 l 3 EI 1 l
l
θC
θC
Construct the Stiffness Equation
∑ M B = 0 →
Pl
∑ M C = 0 → −
Pl
i
i
− θ B = θC =
4 EI 2
2
θC , M CB =
8 8
+(
+
3 EI 1
+
l
2 EI 2 l
4 EI 2 l
θ B + (
)θ B +
3 EI 1 l
+
2 EI 2 l 4 EI 2 l
θC = 0
)θC = 0
Pl 2
1
3 EI 1 + 2 EI 2 8
Member End Forces
M AB = 0 M BA = M BC =
3 EI 1 l Pl 8
+
θ B = −
4 EI 2 l
3 EI 1
Pl
3 EI 1 + 2 EI 2 8
θ B +
2 EI 2 l
θC =
3 EI 1
Pl
3 EI 1 + 2 EI 2 8
Pl 2 EI 2 4 EI 2 3 EI 1 Pl M CB = − + θ B + θC = − 8 l l 3 EI 1 + 2 EI 2 8 M CD =
3 EI 1 l
θC =
3 EI 1
Pl
3 EI 1 + 2 EI 2 8
M DC = 0
In case of EI 1 = 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
20
School of Civil, Urban & Geosystem Eng., SNU 2.3.3 A F rame with an h orizontal force
B
C
P
EI
A
DOF : θ B , ∆
Analysis
i) All fixed : None fixed end moment ii) θ B ≠ 0 1 M AB = 1 M BC =
2 EI l 3 EI l
1
θ B , M BA =
4 EI
l 6 EI
1
θ B , V BA =
l 2
θ B
θ B
iii) ∆ ≠ 0 2 = M AB 2 = V BA
6 EI
l 2 12 EI l 3
6 EI l 2
∆
∆
Construct the stiffness equation
∑ M B = 0 → i
∑V
i
(
4 EI 3 EI 6 EI + )θ B + 2 ∆ = 0 l l l 6 EI
= P →
θ B = −
2
∆ , M BA =
Pl 2 8 EI
2
l
,∆ =
θ B +
12 EI l 3
∆ = P
7 Pl 3 48 EI
Member end forces
M AB =
2 EI l
θ B +
6 EI 2
l
5
∆ = Pl , M BA =
8
4 EI l
θ B +
6 EI 2
l
3
3 EI
8
l
∆ = Pl , M BC =
3
θ B = − Pl
8
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
21
School of Civil, Urban & Geosystem Eng., SNU
2.3.4 A Portal F r ame with an Un symmetri cal L oad
a
P C
B
A
DOF : θ B , θC , ∆
Analysis
D
i) All fixed 0 BC
M
=
Pab 2 l 2
Pa 2b
0 CB
=−
, M
l 2
ii) θ B ≠ 0 1 M AB = 1 M BC =
2 EI l 4 EI l
1
4 EI
1
l 2 EI
θ B , M BA = θ B , M CB =
l
θ B 1
θ B , V BA =
6 EI l 2
θ B
iii) θC ≠ 0 2 M BC = 2 M CD =
2 EI l 4 EI l
2
4 EI
2
l 2 EI
θC , M CB = θC , M DC =
l
θC 2
θC , V CD =
6 EI l 2
θC
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
22
School of Civil, Urban & Geosystem Eng., SNU iv) ∆ ≠ 0 6 EI
3 3 M AB = M BA = 3 3 = V CD = V BA
2
l 12 EI l 3
3
3
∆ , M CD = M DC =
6 EI l 2
∆ ,
∆
Construct the Stiffness Equation
∑ M
i B
=0→
∑ M
i C
∑V
i
Pab 2 l 2
=0→ −
Pa 2b 2
l
+
l 2 EI l
Pa 2b
+
l 2 Pab 2 l 2
l
θ B = −
13 EI
+
2l EI
θ B +
4
EI
84 EI θC =
8 EI l
θ B +
θc +
6 EI 2
l
θc +
6 EI l 2 6 EI l 2
∆=0
∆=0
24 EI l 3
∆=0
θc = 0
2l 13 EI
θ B +
l
θc +
(θ B + θC )
θc = 0
2l 2l 1 Pab ( a + 13b)
∆=
θ B +
2
l
2 EI
θ B +
6 EI
=0→
∆=−
−
8 EI
+
a=
l 4
l 1 Pab (13a + b) 84 EI 1 Pab 28 EI
l (b − a )
2.3.5 A Portal F r ame with a Br acin g (Verti cal Load )
a B
P C
EI = 0, EA ≠ 0
DOF : θ B , θC , ∆
Analysis
A
D
i) All fixed 0 BC
M
=
Pab 2 l 2
0 CB
, M
Pa 2b
=−
l 2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
23
School of Civil, Urban & Geosystem Eng., SNU ii) θ B ≠ 0 1 M AB = 1 M BC =
2 EI l 4 EI
l 6 EI
1 V BA =
l 2
1
4 EI
1
l 2 EI
θ B , M BA = θ B , M CB =
l
θ B θ B
θ B
iii) θC ≠ 0 2 M BC = 2 M CD =
2 EI l 4 EI
l 6 EI
2 V CD =
l 2
2
4 EI
2
l 2 EI
θC , M CB = θC , M DC =
θC
l
θC
θC
iv) ∆ ≠ 0
∆
6 EI
3 3 M AB = M BA = 3 3 M CD = M DC = 3 3 V BA = V CD =
l 2 12 EI l 3
EA ∆
A BD =
2l 2
∆,
l 2 6 EI
∆
2
∆,
∆
1
2l 2
2
= V BD =
EA ∆ 2l 2
Construct the Stiffness Equation
∑ M
i B
=0→
∑ M C = 0 → − i
∑V
i
=0→
Pab 2 2
l Pa 2b l 2 6 EI l 2
+ +
8 EI l 2 EI
θ B +
θ B +
l 6 EI
θ B +
l 2
∆=−
EA ∆
(C) → V BD =
2 EI l 8 EI
θc +
l 24 EI
θc +
1
θc +
l
1+ α 4
l 3
6 EI l 2 6 EI l 2
∆=0
∆=0
(1 + α)∆ = 0
(θ B + θC ) ,
α=
EAl 2 48 2 EI
Solution for b = 3a θ B = −
40 + 52α Pl 2 256(7 + 10α) EI
, θC = −
16 + 28α
Pl 2
256(7 + 10α) EI
, ∆ =
3
Pl 3
128(7 + 10α) EI
For a w × h rectangular section and l = 20h , α = 50 2 . 3 Pl 2 Pl 2 −4 Pl θ B = −0.0203 ∆ = 0.3282 × 10 , θC = 0.0109 EI EI EI
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
24
School of Civil, Urban & Geosystem Eng., SNU
Performance
with Bracing
Response
( α = 50 2 )
w/o bracing (α = 0)
Ratio(%)
θ B (× 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
Μ ΑΒ
( Pl )
-0.0404
-0.0248
162.90
Μ ΒΑ
( Pl )
-0.0810
-0.0694
116.71
Μ CD
( Pl )
0.0438
0.0554
79.06
Μ DC
( Pl )
0.0220
0.0376
58.51
( Pl )
0.1158
0.1216
95.23
A BD ( P )
0.0788
P max ( P all )*
0.0720
0.0685
105.1
P max / vol.
0.0163
0.0228
71.5
Μ P
-
-
*) P all = σ all wh , M all = P all h / 6 Unbalanced shear force in the columns =
6 EI l 2
(θ B + θC ) = 0.0564 P
The bracing carries 99 % of the unbalanced shear force between the two columns.
2.3.6 A Portal F r ame with a Br acin g (H ori zontal L oad)
B
P
C
EI = 0, EA ≠ 0
A
DOF : θ B , θC , ∆
Analysis
D
i) All fixed: No fixed end forces ii)-iv) the same as the previous case
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
25
School of Civil, Urban & Geosystem Eng., SNU
Construct the Stiffness Equation
∑ M B = 0 → i
∑ M C = 0 → i
∑V
i
=0→
8 EI l 2 EI
l 8 EI
θ B +
l 6 EI l 2
2 EI
θ B +
θ B +
l 2
5
5
3
3
θ B = θC , ∆ = − θ Bl = −
θc +
l 6 EI
6 EI
θc +
θc +
l 2 6 EI
∆=0
∆=0
l 2 24 EI l 3
(1 + α) ∆ = P
θC l
Solution θ B = θ C = −
Pl 2
1
(28 + 40α) EI
, ∆=
Pl 3
5
3( 28 + 40α) EI
For α = 50 2 , θ B = θC = −0.3501 × 10
−3
Pl 2 EI
, ∆ = 0.5835 × 10
−3
Pl 3 EI
Performance
Response
with Bracing
w/o bracing (α = 0)
( α = 50 2 )
Ratio(%)
θ B (× 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
0.5835 × 10 −3
0.5952 × 10−1
0.98
0.2801 × 10−2
0.2857
0.98
0.2101 × 10−2
0.2143
0.98
0.2101 × 10−2
0.2143
0.98
0.2801 × 10−2
0.2857
0.98
∆ (× Pl 3 / EI ) Μ ΑΒ
( Pl )
Μ ΒΑ
( Pl )
Μ CD
( Pl )
Μ DC
( Pl )
A BD ( P )
1.4004
-
-
P max( P all)*
0.7141
0.0292
2448
P max / vol.
0.1617
0.0097
1670
*) Governed by A BD for the structure with bracing, and by M DC for the structure without bracing. P all = σall wh , M all = P all h / 6 The bracing carries about 99% of the external horizontal load.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
26
School of Civil, Urban & Geosystem Eng., SNU 2.3.7 A Portal F rame with a Sprin g
a
P C
B
k
A
DOF : θ B , θC , ∆
Analysis
D k ∆
iv) ∆ ≠ 0 3 3 = M BA = M AB 3 3 V BA = V CD =
6 EI
l 2 24 EI 3
l
3
6 EI l 2
∆ ,
3
∆ , V S = k ∆
Construct the Stiffness Equation
∑ M
i B
Pab 2
=0→
∑ M C = 0 → − i
∑V
i
6 EI l 2
3
∆ , M CD = M DC =
l 2 Pa 2b l 2
+ +
=0→
θ B +
8 EI l 2 EI
θ B +
θ B +
l 6 EI l 2
6 EI l 2
θc + (
Deformed Shapes (
24 EI l 3 ∆ S ∆
2 EI l 8 EI
θ B +
θc +
θc +
l 6 EI l 2
θc +
6 EI l 2 6 EI
∆=0
∆=0
l 2 24 EI l 3
∆ = − k ∆
+ k ) ∆ = 0
= 0.41 )
without a spring
EI with a spring ( k = 24 3 ) l
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
27
School of Civil, Urban & Geosystem Eng., SNU 2.3.8 A Portal F r ame Subj ect to Support Settl ement
B
C
A
D δ
DOF : θ B , θC , ∆
Analysis
i) All fixed 0 0 M BC = M CB =
6 EI
δ
l 2 ii)-iv) the same as the previous problem
Construct the Stiffness Equation 6 EI 8 EI 2 EI 6 EI i ∑ M B = 0 → l 2 δ + l θ B + l θc + l 2 ∆ = 0 6 EI 2 EI 8 EI 6 EI i ∑ M C = 0 → l 2 δ + l θ B + l θc + l 2 ∆ = 0 6 EI 6 EI 24 EI i θ B + 2 θc + ∆=0 ∑V = 0 → 2 l l l 3
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
28
School of Civil, Urban & Geosystem Eng., SNU 2.3.9 A Portal F r ame with Un symmetri cal Supports
l /2
C
B
A
DOF : θ B , θC , ∆
Analysis
P
D
i) All fixed 0 M BC =
Pl 8
0 , M CB =−
Pl 8
ii) θ B ≠ 0 1 M BA = 1 M BC = 1 V BA =
3 EI
θ B
l 4 EI
l 3 EI l 2
1
θ B , M CB =
2 EI l
θ B
θ B
iii) θC ≠ 0 2 M BC = 2 M CD = 2 V CD =
2 EI l 4 EI
l 6 EI l 2
2
4 EI
2
l 2 EI
θC , M CB = θC , M DC =
l
θC θC
θC
iv) ∆ ≠ 0 3 M BA =
3 EI l 2
∆ ,
3 3 M CD = M DC = 3 V BA =
3 EI l 3
6 EI l 2 3
∆ ,
∆ , V CD =
12 EI l 3
∆
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
29
School of Civil, Urban & Geosystem Eng., SNU
Construct the Stiffness Equation
∑ M B = 0 → i
∑ M C = 0 → i
∑V
i
Pl 7 EI +
−
8 l Pl 2 EI 8
+
θ B + θ B +
l 3 EI
=0→
l 2
θ B +
2 EI l 8 EI
θc +
l 6 EI l 2
3 EI
θc +
θc +
l 2 6 EI
∆=0
l 2 15 EI l 3
∆=0 ∆=0
l
∆ = − (θ B + 2θC )
5
θ B = −
1 Pl 2 44 EI
, θC =
1 9 Pl 2 44 8 EI
, ∆ = −
1 Pl 3 176 EI
Load Location that Causes No Sidesway
P
a
C
B
A
D l
∆ = − (θ B + 2θC ) = 0 → θ B = −2θC
5
- Stiffness equation
∑ M
i B
=0→
Pab 2 2
l
+
7 EI l
Pab 2 l 2
θ B +
−
2 EI l
12 EI l
θc = 0 ,
θC = 0
Pab 2 l 2
,
∑ M
i C
−
=0→ −
Pa 2b l 2
+
4 EI l
Pa 2b 2
l
+
2 EI l
θ B +
8 EI l
θc = 0
θc = 0
Pa 2b = 3 2 → b = 3a l
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
30
School of Civil, Urban & Geosystem Eng., SNU 2.3.10 A F r ame with a Skewed M ember
P
B
C
A
DOF : θ B , ∆
Analysis
0 = i) All fixed : M BC
Pl Pl 8
+
16
=
3
2 11 0 =− Pl , V BC P 16 2 16
ii) θ B ≠ 0 1 M AB =
2 EI
1 V BC =−
2l
θ B =
3 2 EI l 2
2
2
EI l
1
θ B , M BA = 2 2
EI l
1
θ B , M BC =
3 EI l
EI
1
θ B , V BA = 3
l 2
θ B
θ B
EI 1 2 2 EI 3 θ B =3 θ B l l 2 2 l 2 (2 2
EI l
θ B +
2
EI l
θ B )
1 2l
EI
=3
l 2
θ B
iii) ∆ ≠ 0 2 2 M BA = M AB =
2 = V BC
3 EI 2 l 3
6 EI ∆ 2l 2l
=
3 EI l 2
2
∆ M BC = −
3 EI ∆ l
=−
3 2 EI 2
2l
l 2
2
∆ , V BA = 3 2
EI l 3
∆ ,
∆
3
2 EI 1 1 2 l 2
∆
l 2
=
3 EI 2 l 3
∆
EI EI 1 EI (3 2 ∆ + 3 2 ∆) = 3 2 3 θ B l l l 2l
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
School of Civil, Urban & Geosystem Eng., SNU
31
Construct the stiffness equation
∑ M B = 0 →(2
EI 3 2 EI 3 2 + 3) ) 2 ∆ = − Pl θ B + (3 − l 2 l 16
∑V
3
i
i
= 0 → (3 −
2
2)
EI l 2
θ B + (3 2 +
3 EI 2 11 ) 3 ∆= P 2 l 2 16
EI EI 5.8284 θ B + 0.8787 2 ∆ = −0.1875 Pl l l EI EI 0.8787 2 θ B + 5.7426 3 ∆ = 0.4861 P l l Pl 2
Pl 3 , ∆ = 0.0917 θ B = −0.0460 EI EI
Results
- Deformed shape
- Moment diagram
- Shear force diagram
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
School of Civil, Urban & Geosystem Eng., SNU
32
Chapter 3 Iterative Solution Method & Moment Distribution Method
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
33
School of Civil, Urban & Geosystem Eng., SNU 3.1 Solution Method for Linear Algebraic Equations 3.1.1 Di r ect M ethod – Gauss Eli mi nati on
+ a12 X 2 + L + a1i X i + L + a1n X n = b1 a 21 X 1 + a 22 X 2 + L + a 2i X + L + a 2 n X n = b2 n M → ∑ aij X j = bi a i1 X 1 + ai 2 X 2 + L + aii X i + L + ain X n = bi j =1 M a n1 X 1 + a n 2 X 2 + L + a ni X i + L + a nn X n = bn a11 X 1
for i
= 1L n
or in a matrix form [ A]( X)= (b) By multiplying
a i1 a11
to the first equation and subtracting the resulting equation from the i-th
equation for 2 ≤ i ≤ n , the first unknown X 1 is eliminated from the second equation as follows.
a11 X 1
+
a12 X 2 ( 2) a 22 X 2
+L+ +L+
+ L + a1n X n = b1 a 2( 2i ) X + L + a 2( 2n) X n = b2( 2) a1i X i M
ai(22 ) X 2
+ L + aii( 2) X i + L + ain( 2) X n = bi( 2) M
a n( 22) X 2
where aij( 2)
= aij −
ai1 a1 j
tion by multiplying
a11 ai(22) ( 2) a 22
.
+ L + a n( 22) X i + L + a nn( 2) X n = bn( 2)
Again, the second unknown X 2 is eliminated from the third equa-
to the second equation and subtracting the resulting equation from
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 ( 2) a 22 X 2
+L+ +L+
+ L + a1n X n = b1 a 2( 2i ) X + L + a 2( 2n) X n = b2( 2) a1i X i M
M
a ii(i ) X i
+ L + ain(i ) X n = bi(i ) M (n) a nn X n
M
M
= bn( n )
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
34
School of Civil, Urban & Geosystem Eng., SNU where a
( k ) ij
=a
( k −1) ij
−
ai(,k k −−11) a k ( k −−1,1 j) a
k −1 k −1, k −1
k ≤ i, j ≤ n , and aij1 = a ij .
Once the system matrix is tri-
angularized, the solution of the given system is easily obtained by the back-substitution. bn( n ) (n) ( n) a nn X n = bn → X n = ( n ) a nn a
( n −1) n −1, n −1
X n −1
+a
( n −1) n −1, n
X n
=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
(i ) i
b aii(i ) X i
+ L + ai(,in) X n = bi(i ) →
=
X i
− ∑ aik (i ) X k k = n i −1 ii
a
for 1 ≤ i ≤ n− 1
3.1.2 I terati ve M ethod –Gauss-Jor dan M ethod
A system of linear algebraic equations may be solved by iterative method.
For this purpose,
the given system is rearranged as follows. X 1
=
b1
X 2
=
b2
X i
=
X n
=
− (a12 X 2 + L + a1n X n ) a11
− (a 21 X 1 + a 23 X 3 + L + a 2n X n ) a 22
bi
− (ai1 X 1 + L + ai ,i −1 X i −1 + a i,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
=
1 aii
n
(bi
− ∑ aij ( X j ) k −1 ) j =1 i ≠ j
where the subscript k denotes the iterational count.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
35
School of Civil, Urban & Geosystem Eng., SNU 3.1.3 I terati ve M ethod –Gauss-Siedal M ethod
When we calculate a new X i 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 convergence rate, which leads to the Gauss-Siedal Method. ( X 1 ) k
=
b1
( X 2 ) k
=
b2
( X i ) k
=
( X n ) k
=
− (a12 ( X 2 ) k −1 + L + a1n ( X n ) k −1 ) a11
− (a 21 ( X 1 ) k + a 23 X 3 ) k −1 + L + (a 2n X n ) k −1 ) a 22
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
aii
− ∑ aij ( X j ) k − j =1 i >1
n
∑a
ij
( X j ) k −1 )
j = i +1 i
3.1.4 Exampl e
30
35.6 B
A
4klf C
EI
1.5 EI
D
EI
3@30=90
Stiffness Equation
6 EI 3 EI )θ B + θC = 0 l l l 3 EI 6 EI 3 EI )θC = 0 − 133.3 + 450.0 + θ B + ( + l l l
− 168.7 + 133.3 + (
For the simplicity of derivation,
EI l
3 EI
+
θ B → θ B ,
EI l
θ C → θ C . The stiffness equation becomes
− 35.4 + 9θ B + 3θ C = 0 316.7 + 3θ B + 9θ C = 0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
36
School of Civil, Urban & Geosystem Eng., SNU
Gauss-Jordan Iteration
(θ B ) k
=
(θ C ) k
=
(θ B ) k
=
(θ C ) k
=
1 9 1
(35.4 − 3(θ C ) k −1 )
9
(316.7 − 3(θ B ) k −1 )
Gauss-Siedal Iteration
1 9 1 9
(35.4 − 3(θ C ) k − 1 ) (316.7 − 3(θ B ) k )
Gauss-Seidal
Gauss-Jordan
GAUSS-SIEDAL ITERATION
GAUSS-Jordan ITERATION
======================
======================
***** Iteration 1*****
***** Iteration 1*****
X(1) =
X(1) =
0.3933334E+01
0.3933334E+01
X(2) = -0.3650000E+02
X(2) = -0.3518889E+02
ERROR =
ERROR =
0.1000000E+01
0.1000000E+01
***** Iteration 2*****
***** Iteration 2*****
X(1) =
X(1) =
0.1610000E+02
0.1566296E+02
X(2) = -0.4055556E+02
X(2) = -0.3650000E+02
ERROR =
ERROR =
0.2939146E+00
0.2971564E+00
***** Iteration 3*****
***** Iteration 3*****
X(1) =
X(1) =
0.1745185E+02
0.1610000E+02
X(2) = -0.4100617E+02
X(2) = -0.4040988E+02
ERROR =
ERROR =
0.3197497E-01
0.9044394E-01
***** Iteration 4*****
***** Iteration 4*****
X(1) =
X(1) =
0.1760206E+02
0.1740329E+02
X(2) = -0.4105624E+02
X(2) = -0.4055556E+02
ERROR =
ERROR =
0.3544420E-02
0.2971564E-01
***** Iteration 5*****
***** Iteration 5*****
X(1) =
X(1) =
0.1761875E+02
0.1745185E+02
X(2) = -0.4106181E+02
X(2) = -0.4098999E+02
ERROR =
ERROR =
0.3937214E-03
****** MOMENT ******
0.9812154E-02
****** MOMENT ******
MBA =-115.84375
MBA =-116.34444
MBC = 115.82707
MBC = 115.04115
MCB =-326.81460
MCB =-326.88437
MCD = 326.81458
MCD = 327.03004
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
37
School of Civil, Urban & Geosystem Eng., SNU 3.2 Moment Distribution Method 30
35.6
4klf
B
A
C
EI
D
1.5 EI
EI
3@30=90
At the Joint B
- Moment distribution
=(
M B
3 EI AB L AB
+
4 EI BC L BC
)θ B
1 1 = M BA + M BC → θ B =
M B 3 EI AB 4 EI BC L AB
+
L BC
3 EI AB 1 M BA
=
3 EI AB L AB
θ B =
L AB M 3 EI AB 4 EI BC B
+
L AB
= D BA M B
L BC
4 EI BC 1 M BC =
4 EI BC L BC
θ B =
L BC M 3 EI AB 4 EI BC B
+
L AB 1 - Moment carry over to joint C: M CB
L BC
2 EI BC
=
= D BC M B
L BC
θ B =
1
D BC M B 2
At the Joint C
- Moment distribution M C
=(
4 EI BC L BC
+
3 EI CD LCD
2 2 )θ C = M CB + M CD
→ θ C =
M C 4 EI BC 3 EI CD L BC
+
LCD
4 EI BC 2 M CB
=
4 EI BC L BC
θ C =
L BC M C = DCB M C 4 EI BC 3 EI CD L BC
+
LCD
3 EI CD 2 M CD
=
3 EI CD LCD
θ C =
LCD M C = DCD M C 4 EI BC 3 EI CD L BC
2 - Moment carry over to joint B: M BC
=
+
LCD
2 EI BC L BC
θ C =
1
DCB M C 2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
38
School of Civil, Urban & Geosystem Eng., SNU
Stiffness Equation in terms of Moment at Joints
= 0 M B = → 1 316.7 + D BC M B + M C = 0 M C = 2 − 35.4 + M B +
1
DCB M C 2
1 35.4 − DCB M C 2 1 - 316.7 − D BC M B 2
- Gauss-Siedal Approach 1 DCB ( M C ) k − 1 2 1 ( M C ) k = − 316.7 − D BC ( 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 − D BC ( M B ) k −1 2 ( M B ) k
= 35.4 −
For the given structure
D BA
= DCD =
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 − D BC ( M B )1 = −316.7 − 35.4 → ( ∆ M C )1 2 23 ( M B )1
= 35.4 −
= −328.5
- For k > 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 − D BC ( M B ) k = −316.7 − D BC ( M B ) k −1 − D BC ( ∆ M B ) k 2 2 2 1 1 = ( M C ) k −1 − D BC (∆ M B ) k → (∆ M C ) k = − D BC (∆ M B ) k 2 2 ( M B ) k
= 35.4 −
1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
39
School of Civil, Urban & Geosystem Eng., SNU - Iteration 1 ( M B ) 0 = 0 , ( M C ) 0 = 0 ( M B ) 1
= 35.4 −
1 2
DCB ( M C ) 0
= 35.4 → (∆M B ) 1 =
35.4
f = M BA + D BA (∆ M B )1 = −168.7 + 0.33 × 35.4 = −168.7 + 11.8 → (∆ M BA )1 = 11.8 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
( M BA )1
( M C )1
= −316.7 −
( M CB )1
f = M CB +
1
D BC ( ∆ M B )1 2 1
= −316.7 −
12 23
D BC ( ∆ M B )1 + DCB (∆ M C )1 2
35.4 = −328.5 → (∆M c )1
= − 328.5
= −133.3 + 11.8 − 219.0 → (∆ M CB )1
= 11.8 − 219.0 f = M CD + DCD (∆ M C )1 = 450 − 0.33 × 328.5 = 450 − 109.5 → (∆ M CD )1 = −109.5
( M CD )1 - Iteration 2 (∆ M B ) 2
( ∆ M C ) 2
=−
=−
1 DCB ( ∆ M c )1 2
1 D BC (∆ M B ) 2 2
(∆ M BA ) 2 = D BA (∆ M B ) 2 = 36.5 1 = 109.5 → (∆ M BC ) 2 = DCB (∆ M c )1 + D BC (∆ M B ) 2 2 = −109.5 + 73.0 (∆ M ) = 1 D (∆ M ) + D (∆ M ) CB 2 BC B 2 CB C 2 2 = −36.5 → = 36.5 − 24.3 (∆ 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 M BA
f = M BA + ∑ (∆ M BA ) k = −168.7 + 11.8 + 36.5 + 4.1 = −116.3 k
M BC = M
f BC
+ ∑ (∆ M BC ) k = 133.3 + 23.6 + (−109.5 + 73.0) + (−12.2 + 8.2) = 116.4 k
M CB
= M + ∑ (∆ M CB ) k = −133.3 + (11.8 − 219.0) + (36.5 − 24.3) + (4.1 − 2.7) = −326.9
M CD
= M + ∑ (∆ M CD ) k = 450.0 − 109.5 − 12.2 − 1.4 = 326.9
f CB
k
f CD
k
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
0.33 450.0
-109.5
-12.2
Incremental form for the Gauss-Jordan Method
- For k = 1
= ( M C ) 0 =
( M B ) 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 − D BC ( M B ) 0 = −316.7 → ( ∆ M C )1 = −316.7 2
= 35.4 −
( M B )1
- For k > 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 − D BC ( M B ) k −1 = −316.7 − D BC ( M B ) k − 2 − D BC ( ∆ M B ) k −1 2 2 2 1 1 = ( M C ) k −1 − D BC (∆ M B ) k −1 → (∆ M C ) k = − D BC (∆ M B ) k −1 2 2 ( M B ) k
= 35.4 −
1
- Iteration 1 ( M B ) 0 = 0 , ( M C ) 0 = 0 ( M B ) 1
= 35.4 −
1 2
DCB ( M C ) 0
= 35.4 → (∆M B ) 1 =
35.4
f ( M BA ) 1 = M BA
+ D BA ( M B )1 = −168.7 + 0.33 × 35.4 = −168.7 + 11.8 → (∆ M BA )1 = 11.8 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
41
School of Civil, Urban & Geosystem Eng., SNU ( M C )1
= −316.7 −
1
D BC ( M B ) 0 2
= −316.7 = −316.7 → (∆M c )1 =− 316.7
f = 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
( M CB )1
- Iteration 2 (∆ M B ) 2
( ∆ M C ) 2
(∆ M BA ) 2 = D BA (∆ M B ) 2 = 35.0 1 1 = − DCB (∆ M c )1 = 106.1 → (∆ M BC ) 2 = DCB (∆ M c )1 + D BC (∆ 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 = − D BC (∆ M B )1 = −11.8 → = 11.8 − 7.9 2 (∆ M CD ) 2 = DCD (∆ M C ) 2 = −3.9
- Iteration 3 (∆ M B ) 3
=−
1
( ∆ M C ) 3
=−
1
DCB ( ∆ M C ) 2 2
D BC ( ∆ M B ) 2 2
0.33
(∆ M BA ) 3 = D BA (∆ M B ) 3 = 1.3 1 = 4.0 → (∆ M BC ) 3 = DCB (∆ M C ) 2 + D BC (∆ M B ) 3 2 = −4.0 + 2.7 (∆ M ) = 1 D (∆ M ) + D (∆ M ) CB 3 BC B 3 CB C 3 2 = −35.6 → = 35.6 − 23.9 (∆ M CD ) 3 = DCD (∆ M C ) 3 = −11.7
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
42
School of Civil, Urban & Geosystem Eng., SNU 3.3 Example - MDM for a 4-span Continuous Beam 100 A
1.5 EI , L
EI , L
D BA
DCD
EI
=4 =4
50
L
1.5 EI , L
EI , L
EI 1.5 EI 1.5 EI EI 1.5 EI (4 )= 0.4 , D BC = 4 (4 )= 0.6 , DCB +4 +4 L L L L L
1.5 EI L
(4
1.5 EI L
+4
E
D
C
B
=4
1.5 EI L
(4
1.5 EI L
+4
1.5 EI )= 0.5 , L
1.5 EI 1.5 EI 1.5 EI EI EI 1.5 EI EI )= 0.5 , D DC = 4 (4 (4 + 3 )= 0.67 , D DE = 3 + 3 )= 0.33 L L L L L 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
0.67
0.5 0.0
0.0
81.3 -45.0 22.5
40.6 -90.0 11.3
2.6
7.8 -5.1 2.6
-63.9
64.1
3.9 -10.2 1.3 -0.9 -44.0
22.5 -15.6 7.8
0.33 93.8
-44.3
-5.0 -0.4 44.1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
43
School of Civil, Urban & Geosystem Eng., SNU
Gauss-Jordan Approach 0.4 0.0
0.0 -50
0.5
0.67
0.5
125
-125
-75
62.5
0.0 62.5
0.0 -62.8
31.3
-37.5
-31.4
31.3
-18.8
34.5
34.5
-21.0
17.3
-9.4
-10.5
17.3
-6.9
-10.4
10.0
10.0
-11.6
5.0
-5.2
-5.8
5.0
-2.0
-3.0
5.5
5.5
-3.4
2.8
-1.5
-1.7
2.8
-25.0 -12.5 -6.3 -3.5 -1.0 -1.1
0.33 93.8 -31.0
-10.3 -5.7 -1.6
-1.7
1.6
1.6
-1.9
0.8
-0.9
-1.0
0.8
-0.3
-0.5
1.0
1.0
-0.5
-0.3
-72.8
72.8
-64.4
64.7
-44.0
44.0
-0.6 -36.4
0.6
-0.9
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
43
School of Civil, Urban & Geosystem Eng., SNU
Gauss-Jordan Approach 0.4 0.0
0.6
0.0
125
-50
-125
0.67
0.5
0.0
93.8
-31.0
-75
62.5
62.5
-62.8
-37.5
-31.4
31.3
-12.5
-18.8
34.5
34.5
-21.0
17.3
-9.4
-10.5
17.3
-6.9
-10.4
10.0
10.0
-11.6
5.0
-5.2
-5.8
5.0
-6.3 -3.5 -2.0
0.33
0.0
31.3
-25.0
-10.3 -5.7
-3.0
5.5
5.5
-3.4
2.8
-1.5
-1.7
2.8
-1.1
-1.7
1.6
1.6
-1.9
0.8
-0.9
-1.0
0.8
-0.3
-0.5
1.0
1.0
-0.5
-0.3
-72.8
72.8
-64.4
64.7
-44.0
44.0
-1.0 -0.6 -36.4
0.5
-1.6 -0.9
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
44
School of Civil, Urban & Geosystem Eng., SNU 3.4 Direct Solution Scheme by Partitioning 20 L/3 B
C
A
D
Slope deflection (Stiffness) Equation
8 EI
2 EI
6 EI
44
School of Civil, Urban & Geosystem Eng., SNU 3.4 Direct Solution Scheme by Partitioning 20 L/3 B
C
A
D
Slope deflection (Stiffness) Equation
8 EI L 2 EI L 6 EI L
2 EI
6 EI
L 8 EI
88.8 [K ] θC + − 44.4 = 0 → θθ L (K ∆θ ) 24 EI ∆ 0.0 L
L 6 EI L
L θ B 6 EI
(K θ∆ ) (Θ) K ∆∆
(P) + ∆ 0 = 0
[K θθ ](Θ) = −( P) − (K θ∆ ) ∆ → (Θ) = −[K θθ ] −1 ( P) − [K θθ ] −1 (K θ∆ ) ∆ (K ∆θ )(Θ) P + (K ∆θ )(Θ) ∆
+ K ∆∆ ∆=
= (Θ) P + (Θ ) ∆
0
Direct Solution Procedure by Partitioning
-
Assume ∆ = 0 and calculate (Θ) P .
-
Assume an arbitrary ∆
=
∆=
∆ and calculate ( Θ ) ∆ . α
( Θ) ∆
-
By linearity, ( Θ )
-
Calculate α by the second equation by
α=− -
Obtain (Θ) by
α ∆ and ( Θ ) ∆ .
(K ∆θ )(Θ) P K ∆∆ ∆ + (K ∆θ )(Θ ) ∆
(Θ) = (Θ) P + α( Θ ) ∆ .
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
45
School of Civil, Urban & Geosystem Eng., SNU 3.5 Moment Distribution Method for Frames
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
0.5 C 0 . 5
88.8
-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
33.3
2.1
A
D
M AB = -53.3/2 = -26.7, M DC = -35.5/2 = 17.8, V AB = 2.67, V DC = -1.78, V P = 0.89
Second MDM – with an Arbitrary Sidesway B
C
10.0
10.0
-5.0
-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
1.0
A
-3.8
D
M AB = 10 – 3.9/2.0 = 8.1, M DC = 10 – 4.1/2.0 = 7.9, V AB =-0.47, V DC = -0.46, V ∆ = -0.93
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
46
School of Civil, Urban & Geosystem Eng., SNU
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
47
School of Civil, Urban & Geosystem Eng., SNU
Chapter 4 Energy Principles Pri nciple of M ini mum Potenti al Ener gy and Pri nciple of Vir tual Work
Mg
h
P
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
48
School of Civil, Urban & Geosystem Eng., SNU
Read Chapter 11 (pp.420~ 428) of Elementary Structural Analysis 4 th 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 condi-
tions for rotational angle.
Variables with superscript e denote the exact solution that satisfies
the equilibrium equation(s).
4.1 Spring-Force Systems
∆ P
Total Potential energy
The energy required to return a mechanical system to a reference status ∆
0
Π int = − ∫ k (∆ − u )du = ∫ k (∆ − u )du = ∆
0
Π total = Π int + Π ext =
1 2
1 2
k ∆2 , Π ext = − P ∆
k ∆2 − P ∆
Equilibrium Equation
k ∆e= P
Principle of Minimum Potential Energy for an arbitrary displacement
Π total = = = =
1 2 1 2 1 2 1 2
k ( ∆e
+ ∆ ) 2 − P (∆e + ∆ )
k ( ∆e ) 2
+ k ∆e ∆ +
k ( ∆e ) 2 − P ∆e
+
k ( ∆e ) 2 − P ∆e
+
e = Π total +
∆ = ∆e + ∆ .
1 2
k ( ∆ ) 2
1
k ( ∆ ) 2 − P ( ∆e
2 1 k ( ∆ ) 2 2 1 k ( ∆ ) 2 2
+ ∆)
+ ∆ (k ∆e − P )
e ≥ 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.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
49
School of Civil, Urban & Geosystem Eng., SNU 4.2 Beam Problems q
w
any type of support
l
Π int =
1 M 2
∫
dx
2 0 EI
=
1 2
Π total = Π int + Π ext =
l
l
d 2 w
∫ EI ( dx
= − ∫ qwdx
) dx , Π ext 2
2
0
1 2
l
0
d 2 w
∫ EI ( dx
2
l
∫
) dx − qwdx 2
0
0
Equilibrium Equation
EI
w
Potential Energy of a Beam
any type of support
we
d 2 M e dx 2
= −q
or EI
d 4 w e dx 4
=q
Principle of Minimum Potential Energy for a virtual displacement w = w e
Π = h
1
l
( 2∫
d 2 ( w e dx
0
=
1
l
d 2 w e
( 2 ∫ dx
2
+ w)
EI
2
EI
d 2 w e
0
1 2
l
d 2 w
∫ ( dx
2
EI
d 2 w
0
=Π + e
=Π + e
=Π + e
1 2 1 2 1
l
dx
dx 2
d 2 w
∫ ( dx
2
2
d 2 w
∫ ( dx
2
d 2 w
( 2 ∫ dx 0
2
2
l
∫
) dx − ( w e
+ w )qdx
0
l
∫
) dx − w e qdx + l
∫
)dx + (
EI EI
0
l
dx
+ w)
0
0
l
d 2 ( w e
+w.
EI
0
d 2 w dx 2 d 2 w dx 2 d 2 w dx
2
d 2 w dx 2
EI
d 2 w e dx 2
l
∫
) dx + ( − EI ) dx +
∫ 0
M M e EI
∫
) dx − w qdx 0
d 2 w
0
l
l
1
d 2 w e
l
∫
) ( − EI ) dx − w qdx dx 2 EI dx 2 0 l
∫
dx − w qdx 0
) dx ≥ Π e for all virtual w
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.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
50
School of Civil, Urban & Geosystem Eng., SNU
Principle of Virtual Work
If a beam is in equilibrium, the principle of the virtual work holds for the beam,. l
d 4 w e
∫ w ( EI dx
4
− q)dx = 0 for all virtual displacement
w
0
l
d 2 w
∫ dx
2
EI
d 2 w
∫ dx
2
EI
∫
dx − wqdx +
dx 2
0
l
l
d 2 w e
0
l
d 2 we
0
dx
EI
∫ 0
M M e
∫ 0
EI
l
d 2 w e
− w EI
dx 2
l
d 3 w e
0
l
dx − wqdx =
dx 2
d w
=0
dx 3
0
l
∫
dx − wqdx = 0 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 ~ in a beam by applying an unit load at x ~ yields the displacement of an arbitrary point x and by using the reciprocal theorem. l
l
l
∫ w qdx = ∫ wq dx = ∫ 0
0
wδ( x − x~ ) dx
=
l
M M e
w( x~ ) =
∫
0
0
dx
EI
Approximation using the principle of minimum potential energy
-
Approximation of displacement field w=
n
∑ a g i
i
i =1
-
Total potential energy by the assumed displacement field
Π = h
-
l
l
d 2 w d 2 w ( EI 2 ) dx − wqdx 2 0 dx 2 dx 0
1
∫
∫
1
=
2
l
n
0
i =1
l
n
n
( a g ′′ )dx − ∫ ∑ a g qdx ∫ (∑ a g ′′) EI ∑ i
i
j
j
i
j =1
i
0 i =1
The first-order Necessary Condition l n n ∂Π h ∂ 1 l n ( ( 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
=
1 2
l
∫
l
n
( a g ′′) EI g ′′dx] − ∫ g qdx ) ∑ a g ′′ )dx + ∫ ∑
[ g k ′′ EI ( 0
l
n
j
j
i
i =1
0
i =1
k
k
or
Ka = f
0
l
l
n
i =1 0
0
i =1
n
i
= ∑ ∫ g k ′′ EI g i′′dxai − ∫ g k qdx = ∑ K ki ai − f k = 0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
51
School of Civil, Urban & Geosystem Eng., SNU
Example
P
i) with one unknown
− lx) → w′′ = 2a
w = ax( x − l ) = a ( x 2
Π total =
1 2
l
∫ EI (w′′)
l
2
∫
dx − P δ( x −
0
0
l ) wdx 2
l
l 2 EI ( 2a) dx + aP 20 4
1
=
∫
2
=
l 2 EI 4a l + aP 2 4
1
2
∂Π total l 2 Pl Pl ( x 2 − xl ) = 0 → 4aEIl + P = 0 → a = − →w=− 4 16 EI 16 EI ∂a 3
3
l l w e ( ) − w( ) 2 2 Error = l we ( ) 2
3
l Pl l Pl Pl w( ) = , w e ( ) = , = 0.0208 2 64 EI 2 48 EI EI
= 0.25
ii) with two unknowns w = ax( x − l ) + bx ( x 2
Π total = =
1 2
l
∫ EI (w′′)
l
2
∫
dx − P δ( x −
0
1
0
l 2
EI ( 4a l + 24ab 2 2 2
l ) wdx 2
+ 36b
l 3
2
3
− l 2 ) → w′′ = 2a + 6bx
=
1 2
l
∫ EI (2a + 6bx)
2
dx − P (− a
0
) + P ( a
l 2 4
+b
3l 3 8
l 2 4
−b
3l 3 8
)
)
∂Π total l 2 2 = 0 → EI (4la + 6l b) = − P 1 Pl 4 ∂a a → = − 16 EI ∂Π total 3l 3 2 3 = 0 → EI (6l a + 12l b) = − P 8 ∂b
, b = 0 ???
iii) with three unknowns w = ax ( x − l ) + bx( x 2
Π total =
1
=
1
=
l
− l 2 ) + cx( x 3 − l 3 ) → w′′ = 2a + 6bx+
12cx 2
l
l EI ( w′′) dx − P δ( x − ) wdx 20 2 0 2
∫
∫
2
l
∫ EI (2a + 6bx + 12cx
2
) dx − P (− a 2
0
1
l 3
EI ( 4a l + 36b 2 3
P ( a
2
l 2 4
+b
2
3l 3 8
+c
7l 4 16
+ 144c
2
l 5 5
l 2 4
+ 24ab
−b l 2 2
3l 3 8
−c
+ 48ac
7l 4 16
l 3 3
)
+ 144bc
l 4 4
)+
)
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
52
School of Civil, Urban & Geosystem Eng., SNU
∂Π total l 2 2 3 = 0 → EI (4la + 6l b + 8l c) = − P 4 ∂a ∂Π total 3l 3 2 3 4 = 0 → EI (6l a + 12l b + 18l c) = − P 8 ∂b ∂Π total 144 5 7l 4 3 4 = 0 → EI (8l a + 18l b + l c ) = − P 5 16 ∂c w=
1 Pl 5 P x ( x − l ) − x( x 2 64 EI 32 EI Pl 3
l
1 1
5 3
− l 2 ) +
a = 1 Pl 64 EI 5 P → b = − 32 EI 5 P c = 64 EIl
5 P
x ( x 3 − l 3 ) 64 EIl 21 Pl 3
5 7
Pl 3 0.0205 , Error = 0.0144 EI
w( ) = (− )= + − = 2 EI 64 4 32 8 64 16 1024 EI iv) with one sin function
π
w = a sin x l
Π total =
1
l
l
∫
∫
=→ w′′ = a (
l
EI ( w′′) dx − P δ( x − ) wdx 20 2 0 2
=
1
π
π
) 2 sin x l l
π
l
∫
π
EI (a( ) 2 sin x) 2 dx − aP 20 l l
1 1 π π π l EIa 2 ( ) 4 sin 2 xdx + aP = EIa 2 ( ) 4 + aP 2 l 0 l 2 l 2 l
=
∫
∂Π total 2 Pl 3 2 Pl 3 π 4 l π sin x = 0 → EIa( ) − P = 0 → a = 4 →w= 4 l 2 l ∂a π EI π EI l 1 Pl 3 Pl 3 e l w( )= , w ( ) = , 2 48.7045 EI 2 48 EI
Error = 0.0145
v) with two sin function 3π π w = a sin x + b sin x l l
Π total =
1
=
1
=
1
l
=→ w′′ = a(
π
3π 3π π ) 2 sin x + b( ) 2 sin x l l l l
l
l EI ( w′′) dx − P δ( x − ) wdx 20 2 0
∫
∫
2
3π 3π π π EI ( a( ) 2 sin x + b( ) 2 sin x ) 2 dx − aP + bP 20 l l l l l
∫
2
π
l
EIa ( ) 2 l
4
∫ 0
sin
2
π
π
3π
xdx + EIab( ) ( ) l l l 2
l
2
∫ 0
sin
π l
sin
3π l
xdx +
3π 3π EIb 2 ( ) 4 sin 2 xdx − aP + bP 2 l 0 l l
1
=
∫
3π l π l 1 EIa 2 ( ) 4 + EIb 2 ( ) 4 − aP + bP 2 l 2 2 l 2
1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
53
School of Civil, Urban & Geosystem Eng., SNU
∂Π total 2 Pl 3 π 4 l = 0 → EIa( ) − P = 0 → a = 4 ∂a l 2 π EI ∂Π total 3π 4 l 2 Pl 3 = 0 → EIb( ) + P = 0 → b = − l 2 ∂b (3π) 4 EI 2 Pl 3 1 3π π w= 4 (sin x − sin x) l 81 l π EI l Pl 3 Pl 3 Pl 3 e l w( ) = 0.0205(1 + 0.0123) , w ( )= 0.0208 , Error ≅0 = 0.0208 2 EI EI 2 EI
4.3 Truss problems
− F ji − F mi (i ) X i
X i
Y i
− F 1i
Y i
Potential Energy
Π int =
2 1 nmb ( F i ) l i
2
∑ i =1
njn
,
EA i
Π ext = −∑ ( X i u i + Y i vi ) i =1
Π total = Π int + Π ext =
2
1 nmb ( F i ) l i 2
∑ i =1
EA i
njn
− ∑ ( X i u i + Y i vi ) i =1
where nmb and njn denotes the total number of members and the total numbers of joints in a truss.
Equilibrium Equations m (i )
− ∑ H + X = 0 i j
i
j =1
m (i )
,
−∑
V ji
+ Y i = 0
for
i
= 1,L, njn
j =1
where m(i), H ji 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.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
54
School of Civil, Urban & Geosystem Eng., SNU
Principle of Minimum Potential Energy
Π total = = =
1 nmb ( F i
+ F i
2
EAi
e
∑ i =1
e 2 1 nmb ( F i ) l i
2
∑ ∑
=Π + e
i =1
i =1
∑
i =1
2
njn
nmb
( F i ) 2 l i
i =1
i =1
EAi
2 1 nmb ( F i ) l i
∑ 2
2 1 nmb ( F i ) l i i =1
− ∑ ( X i u i + Y i vi ) + ∑
EAi
i =1
− ∑ ( X i (u i +u i ) + Y i (vi + vi ))
− ∑ ( X i u i + Y i vi ) +
e 2 1 nmb ( F i ) l i
2
njn
njn
EAi
i =1
) 2 l i
EAi
≥ Πe
EAi
nmb
F i e F i l i
i =1
EAi
+∑
njn
− ∑ ( X i u i + Y i vi ) i =1
for all virtual displacement fields
e
where F i and F i 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.
Virtual Work Expression
If a truss is in equilibrium, the principle of the virtual work holds for the truss,. njn
m (i )
∑ ((−∑ H + X )u i j
i =1
j =1
i
m (i )
i
+ (-∑ V ji + Y i )v i ) = 0 j =1
(v 2
(u 2
− v1 ) sinθ i (v 2 − v1 )
− u1 ) cosθ i
(u 2 − u1 )
θi
θi
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
55
School of Civil, Urban & Geosystem Eng., SNU njn
m (i )
∑ ((−∑ F cos θ i j
i =1
m (i )
j
j =1
njn
j =1
m (i )
∑ (u ∑ F cos θ i
i j
i =1
+ X i )u i + (-∑ F ji sin θ j + Y i )vi ) = 0
j
+ δv
j =1
nmb
∑ ( F cos θ (u e
i
i
2 i
m (i ) i
njn
∑ F sin θ ) = ∑ ( X u i j
j
i
j =1
− u ) + F i 1 i
e
sin θ i (v
2 i
−v
1 i
)) =
i =1
nmb
∑ F (cos θ (u e
i
i
2 i
F i e F i l i
i =1
( EAi )
∑
+ Y i vi )
njn
∑ ( X u i
i
+ Y i vi )
i =1
− u ) + sin θ i (v − v 1 i
2 i
i =1
nmb
i
i =1
1 i
)) =
nmb
nmb
∑ F ∆l = ∑ F e
i
i =1
e
i
i
i =1
F i l i ( EAi )
nmb
F i e F i l i
i =1
( EAi )
=∑
n
= ∑ ( X i u i + Y i vi ) 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. X k u k + Y k v k Since
α represnts
=
X u k cos α = u k cos α =
nmb
F i e F i l i
i =1
( EAi )
∑
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.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
School of Civil, Urban & Geosystem Eng., SNU
56
Chapter 5 Matrix Structural Analysis
Mr. Force & Ms. Displacement
Matchmaker : Stiffness Matrix
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
57
School of Civil, Urban & Geosystem Eng., SNU 5.1 Truss Problems 5.1.1 M ember Stif fn ess M atr ix
f y L
f x L
δ R y
f y R
δ L y
δ L x
f x R
δ R x
Force – Displacement relation at Member ends
EA
f x L
=−
f x R
=
f y L
= f y R = 0
L EA L
L (δ R x − δ x )
(δ R x
− δ L x )
Member Stiffness Matrix in Local Coordinate System e
f x L 1 L f y EA 0 f R = L − 1 x f y R 0
0
−1
0
0
0
1
0
0
0 δ L x
e
L 0 δ y 0 δ R x R 0 δ y
(f ) e = [k ]e ( ) e
Transformation Matrix
V y
v y
v x
è ¨
V x
φ ç ¨ V x V y
= cos φv x − sin φv y V x → = 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 )
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
58
School of Civil, Urban & Geosystem Eng., SNU
Member End Force e
F x1 cos φ − sin φ 1 F y sin φ cos φ F 2 = 0 0 x2 F 0 0 y
0
Member End Displacement e
δ L x cos φ L δ y − sin φ δ R = 0 R x δ 0 y
e
L f x 0 0 f y L R → (F)e = [Γ ]T (f )e cos φ − sin φ f x R sin φ cos φ f y
0
e
1 ∆ x 1 cos φ 0 0 ∆ y 2 → (δ)e = [Γ](∆ )e 0 cos φ sin φ ∆ x 2 0 − sin φ cos φ ∆ y
sin φ
0
0
Member Stiffness Matrix in Global Coordinate
(F) e
= [Γ]T (f )e = [Γ]T [k ]e (
)e
= [Γ]T [k ]e [Γ](
)e
(F) e = [K ]e ( ) e
[K ]
e
cos 2 φ sin θ cos φ − cos 2 φ − sin φ cos φ sin 2 φ − sin φ cos φ − sin 2 φ [K 11 ]e EA sin θ cos φ = = e L cos 2 φ sin φ cos φ [K 21 ] − cos 2 φ − sin θ cos φ sin 2 φ − sin 2 φ sin φ cos φ − sin φ cos φ
[K 12 ]e
[K 22 ] e
5.1.2 Gl obal Stif fn ess Equati on
− (F1 ) m (3)
Nodal Equilibrium
− (F 2 ) m ( 2)
− (F 2 ) m ( 4 )
n-th joint
− (F1 ) m(1)
− (F1 ) m(5) Pm
Pm
m-th joint 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
59
School of Civil, Urban & Geosystem Eng., SNU
Pn
n-th joint
(F 2 ) i 2
(F1 ) i
1 i-th member
P
m
m-th joint
0 0 nm(1) 1 or 2 1(k ) F ( ) M M (P)1 ∑ k =1 1 i M F ( ) I M nm ( m ) p p ( P) = ( P) m = ∑ (F1 or 2 ) m ( k ) = ∑ M = ∑ M = i =1 (F 2 )i i =1 0 M k 1 M (P)q nm ( q ) M (F1 or 2 )q ( k ) M ∑ 0 0 k =1
[E] = [[E]1 L [E]i L
0
m-th row 0 (F1 )i p i i M 2 i =∑ [E] (F ) = (F ) i =1 I n-th row M 0 M
[E](F)
(F)1 M [E] p ] , (F) = (F) i M (F) p
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
60
School of Civil, Urban & Geosystem Eng., SNU
Compatability Condition un
n
(
2 i
)
2
m
1
um
(
1 i
)
i-th member
1 i
(
)
= um ,
(
2 i
)
= un → (
)
i
( = (
I = 2 i ) 0
1 i
)
m-th column
( )
i
( = (
0 = 2 i ) 0
1 i
)
L
I
L
0 L
L
0 L I L
n-th column
(
0 u m
I u n
u1 M u m 0 M = [C]i (u) 0 un M u q
( )1 [C]1 u1 ) = M = M M = [C](u) ( ) p [C] p u q Compatibility Matrix
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]
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
61
School of Civil, Urban & Geosystem Geosys tem Eng., SNU
Unassembled Member Stiffness Equation
(F)1 [K ]1 M M (F ) i = 0 M M p (F) 0
L
0
L
L
M
L
L [K ]
i
L
L
M
L
L
0
L
(∆)1 M 0 ( ∆) i → M M [K ] p ( ∆) p 0
(F) = [K ](∆ )
Global Stiffness Equation
( P) = [C]T (F) = [C]T [K ](∆ ) = [C]T [K ][C](u) (P )= [K ](u)
where
[K ] = [C]T [K ][C]
Direct Stiffness Method
[C]1 M i 0 [C] M M p [K ] p [C]
[K ]1 L 0 L M L M L T 1 i p i [K ] = [C] [K ][C] = [[C] L[C] L[C] ] 0 L [K ] L M L M L 0 L 0 L = [C]1 [K ]1 [C]1 + L + [C]i [K ]i [C]i + L + [C] p [K ] p [C] p T
T
i T
i
[C] [K ] [C]
i
T
T
T
0 M I = M M M 0 0 M I = M M M 0
0
T
0
0 [K 11 ]i M [K 21 ]i I M 0 0 M 0 0 L M 0 L I M 0 M
[K 12 ]i 0 L I
[K 22 ]i 0
L
L
0 L 0
0 L I
L
m-th row
[K 11 ]i
L
[K 12 ]i
[K 21 ]i
L
[K 22 ]i
0
0 0 L 0 = 0 L 0 0 n-th row 0
0 [K 11 ]i 0 [K 21 ]i 0
m-th column
0
0
0 [K 12 ]i 0
0
0 [K 22 ]i 0
0
0
0 0 0 0
n-th column
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
62
School of Civil, Urban & Geosystem Geosys tem Eng., SNU 5.1.3 5.1.3 Exampl e
P 4, u4
P 3, u3 2
L
2
1
P 2, u2
1
P 1, u1
P 6, u6
3
P 5, u5
3 2 L
Member Stiffness Matrix
[K ]
e
cos 2 φ sin φ cos φ − cos 2 φ − sin φ cos φ sin 2 φ − sin φ cos φ − sin 2 φ [K 11 ]e EA sin φ cos φ = = e L cos 2 φ sin φ cos φ [K 21 ] − cos 2 φ − sin φ cos φ sin 2 φ − sin 2 φ sin φ cos φ − sin φ cos φ
[K 12 ]e
[K 22 ] e
- Member 1: φ = 45o 1 1 1 1 − − 2 2 2 2 1 1 1 1 − − EA 1 2 2 2 2 [K ] = 1 1 1 1 2 L − − 2 2 2 2 1 1 1 1 − − 2 2 2 2 - Member 2: φ =-45 o
[K ] 2
1 1 −1 −1 2 2 2 2 1 1 1 1 − − EA 2 2 2 2 = 1 1 1 1 2 L − − 2 2 2 2 1 1 1 1 − − 2 2 2 2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
63
School of Civil, Urban & Geosystem Geosys tem Eng., SNU - Member 3: φ = 0o
[K ]
3
1 EA 0 = 2 L − 1 0
Equilibrium Equation
0
−1
0
0
0
0
0
1
0
0
0 0
P 4 P 3
P 2
P 6
P 5
P 1
P 1 = ( F x1 )1
+ ( F x2 ) 3 P 2 = ( F y1 )1 + ( F y2 ) 3 = ( F x2 )1 + ( F x1 ) 2 P 4 = ( F y2 )1 + ( F y1 ) 2 P 3
= ( F x2 ) 2 + ( F x1 ) 3 P 6 = ( F y2 ) 2 + ( F y1 ) 3 P 5
P 1 1 P 2 0 P 0 3= P 4 0 P 0 5 P 6 0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
1
0
1
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
1
0 1
0
[E]1
[E]2
[E]3
F 1 1 x F y1 2 F x 2 0 F y 2 1 F x1 1 0 F y 0 F x2 0 F y2 0 F 1 3 x F 1 y2 F x F 2 y
( P) = [[E1 ] [E 2 ] [E 2 ]](F) = [E](F )
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
64
School of Civil, Urban & Geosystem Eng., SNU
Compatibility Condition
u4
u3
u2
u6
u5
u1
∆1 1 x 1 ∆1 y 0 2 ∆ x 0 ∆2 y 2 0 ∆1 x 0 1 ∆ y 0 ∆2 = 0 2 x ∆ y 0 1 3 0 ∆ x ∆1 0 2 y ∆ x 1 ∆2 0 y
0
0
0
0
1
0
0
0
0 1
0
0
0
0
1
0
0
1
0
0
0
0
1
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
[C]1 0 0 u1 u1 0 u 2 u [C]1 2 0 u 3 = [C]2 u 3 u 0 u 4 [C]3 4 1 u 5 u5 u 0 u 6 6 1 [C]2 0 [C]3 0 0
( ) = [C](u) [E1 ] = [C1 ]T , [E 2 ] = [C 2 ]T , [E 3 ] = [C 3 ]T
→
[E]= [C]T
Global Stiffness Matrix T
[K ] = [C]T [K ][C] = [C]1 [K ]1 [C]1
+ [C] 2
T
[K ] 2 [C] 2
T
+ L + [C]3
[K ]3 [C]3
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
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 1 0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0 1 0 2 1 0 2 1 1
1 2 1 2 1
0
0
1
0
0 1 0
0
0
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
0
0
0
0
1
0
0
1
0
0
− −
0 0
− −
0
2 1
2 1 2 1
1
−
0
2 1
−
0
2 1
2 1
2 1
2
2
0 0
−
1 2 1 2 1
1
−
1 2 1 2 1
2 1 2 1 2 1
0 0 0 0 0 0
0
0 0 0 0 0
1
2 0 1 − 0 2 1 0
− − 2 2 2 2 0 0 1 1 −1 −1 1 2 2 2 2 0 0 1 1 1 − − 2 2 2 0 1 1 1 − EA 0 2 2 2 = 1 1 1 2 L − 2 2 2 0 1 1 1 − − 2 2 2 0
2
0 1
1
1 − 2 2 1 0 0 0 0 0 1 1 − − 0 1 0 0 0 0 2 2 1 1 0 0 1 0 0 0 2 2 0 0 0 1 0 0 1 1 2 2 1 1 1 1 − − 2 2 2 2 0 1 1 1 1 − − 2 2 2 2 0 1 1 1 1 = EA − 2 −2 2 2 2 L 0 1 1 1 1 − − 2 2 2 2 0 0 0 0 0 0 0 0 0
−
2 1
0 1 0 2 1 0 − 2 1 0
1
0
1
− − 2 0 2 1 −1 0 − 2 2
− − 2 2 0 1 1 − 0 − 2 2
0 0 EA 1 [C 2 ]T [K 2 ][C 2 ] = 2 L 0 0 0 1 0 EA 0 = 2 L 0 0 0
0 1 0 2 1 0 2 1 1
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0 1 0
0
0
0
0
0
0
0 1
0 1
0 1
0 1
2 1
2 1
0 0
−
0
−
0
2 1
−
2 1
−
2 1
2 1
2 1
2 1
2 1
2 1
−
2
2
−
2
− 2 1 − 2 1 2
0
1 0 0 0 − 1 0 0 0
0
0 1
0
−1
0
0
0
0
0
0
0
1
0
0
0
0
1
0
0 0
0 0 0 1 0 0
0
0
0 1
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0 0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
66
School of Civil, Urban & Geosystem Eng., SNU
0 0 EA 0 = 2 L 0 1 0
0 1
0
0
0
1
0
0
0
0
0
0
1
0
− 1 0 0 0 1 0 0 0
0
0
0
1
0
0
0
0
0
0
0
−1
0
0
0
0
1 0 0 0 EA 0 = 0 2 L 0 − 1 0 0
1 1 1 1 0 0 − − 2 2 2 2 1 1 1 1 0 0 − − 2 2 2 2 1 1 1 EA 1 0 0 + − − [K ] = 2 2 2 2 2 L 1 1 1 1 0 0 − − 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 − 1 0 0 0 0 0 0 0 0 0 EA 0 0 0 0 + 0 0 2 L 0 0 0 0 − 1 0 0 0 1 0 0 0 0 0 0 0 1 1 1 +1 − 2 2 2 2 2 2 2 1 1 1 − 2 2 2 2 2 2 1 1 1 1 − − + EA 2 2 2 2 2 2 2 2 = 1 1 1 L − 1 − − 2 2 2 2 2 2 2 2 1 −1 0 − 2 2 2 1 0 0 2 2
0 0 0 EA 0 2 L 0 0
− −
0
0
0
−1
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
0
0
0
0 1
0 1
0 1
0 1
2 1
2 1
0 0
−
0
−
0
2 2
+
2 2 1 2 2
1 2 2 1
−
2 2
−
2 1
2 1
2 1
2 1
2 1
2 1
2 1
−
2
−
2 2 1
2 2 1 1 2 2 1
−
2 1
1
−
0 0 0 0
−
2
1 2
1 2 2 1
2 2 1 1 2 2
−
+
1 2 2
− 2 1 − 2 1 2
0 1 2 2 1 − 2 2 1 − 2 2 1 2 2 0
0
−
2
2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
67
School of Civil, Urban & Geosystem Eng., SNU
Stiffness Equation
Unknown Unknown Known Known Known Unknown
1 + 2 1 1 1 1 0 − − − 2 2 2 2 2 2 2 2 2 1 1 1 1 0 0 u1 − − P 1 2 2 2 2 2 2 2 2 P 2 1 1 1 1 1 u 2 0 − − P EA − 3 2 2 2 2 2 2 2 2 2 u 3 = 1 1 1 u 4 P 4 L − 1 − 1 0 − 2 2 P 2 2 2 2 2 2 2 u 5 5 1 1 1+ 2 1 P −1 0 − − u 6 6 2 2 2 2 2 2 2 2 2 1 1 1 1 0 0 − − 2 2 2 2 2 2 2 2
Known Known Unknown Unknown Unknown Known
Application of Support Conditions (Boundary Conditions)
1 + 2 1 1 1 1 0 − − − 2 2 2 2 2 2 2 2 2 1 1 1 1 0 0 u1 − − P 1 2 2 2 2 2 2 2 2 P 2 1 1 1 1 1 u 2 u 0 − − P EA − 3 2 2 2 2 2 2 2 3 2 2 = 1 1 1 u 4 P 4 L − 1 − 1 0 − 2 2 P 2 2 2 2 2 2 2 u 5 5 1 1 1+ 2 1 P −1 0 − − 6 u 6 2 2 2 2 2 2 2 2 2 1 1 1 1 0 0 − − 2 2 2 2 2 2 2 2
Final Stiffness Equation
1 P 3 2 EA P 4 = 0 L P 5 − 1 2 2
0 1 2 1 2 2
1 − 2 2 u 3 u 3 2 1 L u4 → u4 = 0 EA 2 2 u u 5 5 1+ 2 − 1 2 2 2 2 1
0 1 2 1 2 2
− 2 2 1 2 2 1+ 2 2 2
−1
1
P 3 P 4 P 5
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
68
School of Civil, Urban & Geosystem Eng., SNU 5.2 Beam Problems 5.2.1 M ember Stif fn ess M atri x
m R
m L
=
4 EI e
m R
=
2 EI e
=
f
Le Le
θ L +
2 EI e
θ L +
4 EI e
M Le + M R e Le
Le Le
=
M Le + M R e
=−
R y
f
Le
θ R +
6 EI e
θ R +
6 EI e
6 EI e 2 e
L
=−
2 e
L
2 e
L
6 EI e
δ L y −
6 EI e
2 e
L
θ L +
6 EI e
δ L y −
6 EI e 2 e
L
θ L −
2 e
L
2 e
L
θ R +
6 EI e 2 e
L
δ R y δ R y
12 EI e 3 e
L
θ R −
δ L y −
12 EI e 3 e
L
12 EI e 3 e
L
δ L y +
δ R y
12 EI e 3 e
L
δ R y
Transformation Matrix is not required
f → F , m → M ,
θ R
Force-Displacement Relation at Member Ends
L y
L y
θ L
m L
δ R y
f y R
δ
f y L
δ → ∆ , θ →Θ
Member Stiffness Matrix e
F y1 1 M = EI e L e 2 F y 2 M
6 12 12 − L2 Le L2e e 6 6 4 − Le Le 12 6 12 − 2 − Le L2e Le 6 6 2 − L Le e
1 6 ∆ y Le 1 2 Θ 6 − ∆2 Le y 4 Θ y2
e
or (F) e = [K ]e ( ) e
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
69
School of Civil, Urban & Geosystem Eng., SNU 5.1.2 Global Sti ff ness M atri x
Nodal Equilibrium
Vi M i
i-th Member
(i-1)-th Member
= ( F y2 ) i −1 + ( F y1 ) i i R P F ( ) → = i −1 + M i = ( M 2 ) i −1 + ( M 1 ) i V i
P1 (F) L1 R L 2 P (F )1 + (F ) 2 M M R L i − 1 P (F ) + ( F ) i −1 i−2 P i = (F) Ri−1 + (F) Li = [[E]1 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 0 0 M M (F) L I Ri = ( F ) i 0 M M 0 0
i-th row (i+1)-th row
Compatibility
= u2i −1
(Θ1 )i
= u2 i
(∆2 y )i
= u2i +1
(Θ 2 ) i
= u2 i + 2
→
(F)1 M [E] p ] (F) i = [E](F) M (F) p
L
0
0 (F ) Li = [E]i (F)i I (F) Ri M 0 M
u2i
u2i − 1
1
u 2i
u2i + 2
i-th Member
(i-1)-th Member (∆1 y )i
L [E]i
(F) Li
L i
( )
=u
i
R i
, ( )
i +1
= u →(
)
i
( = (
) Li
I = ) Ri 0
0 u i
I u i +1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
70
School of Civil, Urban & Geosystem Eng., SNU i-th column
( )i
u1 M L i ) i 0 L I 0 L 0 u = = R i +1 ) i 0 L 0 I L 0 u M (i+1)-th column p +1 u ( )1 [C]1 u1 ) = M = M M = [C](u) ( ) [C] u p +1 p p
( = (
(
Unassembled Member Stiffness Equation
(F)1 [K ]1 M M (F) = 0 i M M ( ) F p 0
[C]i (u)
L
0
L
L
M
L
L [K ]i
L
L
M
L
L
0
L
(∆)1 M 0 (∆) i → M M [K ] p ( ∆) p 0
(F) = [K ](∆ )
Global Stiffness Equation
( P) = [C]T (F) = [C]T [K ](∆ ) = [C]T [K ][C](u) (P )= [K ](u)
where
[K ] = [C]T [K ][C]
Direct Stiffness Method
T i
[C] [K ]i [C]i
0 M 0 = 0 M 0
0
0
0
0
0
M L
M
M
M
[K 11 ]i
[K 12 ]i
M L
L [K 21 ]i
[K 22 ]i
L
M
M
M
M
0
0
0
0
i-th column
0 0 M 0
i-th row i+1-th row
i+1-th column
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
71
School of Civil, Urban & Geosystem Eng., SNU 5.2.3 Exampl e
V 1 , w1 M 1 , θ1
V 3 , w 3
V 2 , w 2
V 4 , w 4 M 3 , θ 3
M 2 , θ 2
M 4 , θ 4
Equilibrium Equation
, M 1 = M 1 L = V 1 L V 2 = V 1 R + V 2 L , M 2 = M 1 R + M L2 L V 3 = V 2 R + V 3 L , M 3 = M R 2 + M 3 V 4 = V 4 R , M 4 = M R 4 V 1
V 1 M 1 2 V M 2 (P ) = 3 V 3 M V 4 M 4
1 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
0
0
0 1
0 1
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 1
0
1
0
0
0
0
0
0
0
0
0
0 1
0
0
0
0
0
0
0
0
0
[E]1
[E]2
0
[E]3
V 1 L L M 1 R V 0 1 R 0 M 1 L 0V 2 L 0 M 2 = 0V 2 R 0 M R 2 0V L 3 1 L M 3 V R 3 M R 3
[E](F)
Compatability Condition
w L 1 w L2 w L3
= w1 , θ L1 = θ1 , w R1 = w 2 , θ R1 = θ 2 = w 2 , θ L2 = θ 2 , w R2 = w 3 , θ R2 = θ 3 = w 3 , θ L3 = θ 3 , w R3 = w 4 , θ R3 = θ 4
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
72
School of Civil, Urban & Geosystem Eng., SNU
w L1 1 L θ1 R 0 w1 0 θ R 1 0 w L2 0 L θ 2 0 (∆) = R = w2 0 θ R 0 2 w L3 0 L 0 θ3 w R 0 3 0 θ R 3
0
0
0
0
0
1
0
0
0
0
0
0 1
0
0
0
0
0
0
1
0
0
0
0 1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0 1
0
0
0
0
1
0
0
0
0
0
0 1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0 1 0 w 1 0 θ 2 0 w 0 θ 2 0 w3 0 θ 3 0 w 4 0 4 θ 0 1
[C]1
w1 θ1 2 w [ ] C 1 2 = [C]2 θ w 3 [C]3 3 θ w4 θ4 [C]3
=
[C](u)
[C]2
Unassembled Member Stiffness Matrix
(F)1 [K ]1 ( F ) = (F ) 2 = 0 (F ) 0 3
0
0 [K ] 2 0
(∆)1 0 ( ∆) 2 = [K ]3 (∆) 3 0
[K ](∆)
Global Stiffness Equation
( P) = [E](F) = [E][K ](u) = [C]T [ K ][C](u)= [K ](u)
[C]1 [C] (u) ( P) = [[C]1T , [C]T 2 [K ] 2 0 2 0 [K ]3 [C]3 = [K ](u) = ( [C]1T [K ]1[C]1 + [C]T 2 [K ]2 [C]2 + [C]T 3 [K ]3 [C] 3 )(u) [K ]1 , [C]T 3 ] 0 0
0
0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
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 0 2 0 L 0 6 1 L1 0 12 − 0 L2 1 0 6 0 L1
0 0 1 0 0 0 0 0
6 12 L2 L1 1 6 4 L1 12 6 − − EI 1 L2 L1 1 L1 6 2 L 1 0 0 0 0 0 0 0 0 -
− −
12 L12 6
L1 12
L12 6
−
L1 0 0 0 0
L1 L L1 6 1 4 − 2 0 L1 6 12 6 0 − − 0 L1 L12 L1 6 2 − 4 L1 6 0 0 0 0 L1 2 0 0 0 0 6 0 0 0 0 L1 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6
12
−
6
2 1
0 0 1 0 0 1 0 0
0 0 0 1
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 0
0 0 0 0 1 0 0 1
0 0 0 0
0 0 0 0
=
Member 2
EI 2 L2
EI 2 L2
0 0 1 0 0 0 0 0
0 0 0 0 0 0 1 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 12 2 0 L2 0 6 0 L2 0 12 − 1 L2 2 0 6 0 L
0 0 6
L22 6
L2
−
2 2
L 6
L2 0 0
L2 4
−
4
−
6
−
L2 2
− −
−
12 2 2
L 6
L2 12 L22 6 L2
0 0 12
0 0 6
L22 6
L2
L2 12 2 2
L2
L
2
−
0 0
−
6
2
0 0 12
L2 12
6
6 L2 0 0
2
−
6 L2 4 0 0
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 6
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
74
School of Civil, Urban & Geosystem Eng., SNU -
Member 3 0 12 2 0 L3 0 6 0 L3 0 12 − 0 L2 3 0 6 1 L
0 0 0 EI 3 0 L3 1 0 0 0
0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0
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
6 L3 4
−
− −
6 L3 2
−
3
0 0 0 0 12
0 0 0 0 6
L23 6
L3
L3 12
−
L23 6
L3
4
−
6 L3 2
− −
12 2 3
L 6
L3 12 L23 6 L3 0 0 0 0 12 L23 6
L3 12
L23 6
−
L3
L3 2 0 0 6 0 − L3 0 4 0 0 0 0 6 L3 2 6 − L3 4 6
0 0 0 0
0 0 0 0
0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 1 0 0 0 1
=
Global Stiffness Matrix
12 I 1 6 I 1 12 I 1 6 I 1 0 0 0 0 − 2 3 2 L3 L1 L1 L1 1 6 I 1 2 I 1 6 I 1 4 I 1 0 0 0 0 − 2 2 L1 L1 L1 L1 12 I 6 I 12 I 12 I 6 I 6 I 12 I 2 6 I 2 1 2 1 2 − 3 1 − 21 0 0 + 3 − 2 + 2 − 3 3 2 L1 L1 L2 L1 L2 L2 L2 L1 6 I 1 2 I 1 6 I 6 I 4 I 1 4 I 2 6 I 2 I 2 0 0 − 21 + 22 + − 22 2 L1 L1 L1 L2 L2 L2 L1 L2 E 12 I 2 6 I 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 6 I 3 2 I 3 6 I 2 2 I 2 6 I 2 6 I 3 4 I 2 4 I 3 0 0 − 2 + 2 + − 2 L22 L2 L2 L3 L2 L3 L3 L3 12 I 3 6 I 3 12 I 3 6 I 3 0 0 0 0 − 3 − 2 − 2 3 L3 L3 L3 L3 6 I 3 2 I 3 6 I 3 4 I 3 0 0 0 − 2 0 2 L L L L 3 3 3 3
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
75
School of Civil, Urban & Geosystem Eng., SNU
V 1 M 1 2 V M 2 V 3 3 M V 4 M 4
Application of support Conditions
12 I 1 6 I 1 12 I 1 6 I 1 0 0 0 0 − L3 L12 L13 L12 1 4 I 1 6 I 1 2 I 1 6 I 1 0 0 0 0 − 2 2 L1 w1 L1 L1 L1 12 I 6 I 12 I 12 I 6 I 1 6 I 2 12 I 2 6 I 2 1 1 1 2 0 0 θ1 − 3 − 2 + 3 − 2 + 2 − 3 3 2 L1 L1 L2 L1 L2 L2 L2 L1 2 w 6 I 1 2 I 1 6 I 1 6 I 2 4 I 1 4 I 2 6 I 2 2 I 2 0 0 2 − 2 + 2 + − 2 2 L L L L L L L L 1 1 2 2 2 1 2 θ = E 1 12 I 3 6 I 3 w 3 12 I 2 6 I 2 12 I 2 12 I 3 6 I 2 6 I 3 0 0 − − + − + − L32 L22 L32 L33 L22 L23 L33 L23 3 θ 6 I 4 I 6 I 2 I 6 I 2 I 6 I 4 I 3 3 3 3 0 2 2 2 2 0 − + + − w4 2 2 2 2 L2 L2 L2 L3 L2 L3 L3 L3 4 12 I 3 6 I 3 12 I 3 6 I 3 θ 0 0 0 0 − 3 − 2 − 2 3 L3 L3 L3 L3 6 I 3 2 I 3 6 I 3 4 I 3 − 2 0 0 0 0 L23 L3 L3 L3
Final Stiffness Equation
4 I 1 4 I 2 L + L 2 2 M 2 2 I 2 3 M L2 E = 4 V 0 M 4 0
2 I 2
0
L2 4 I 2 L2
−
+
4 I 3
L3 6 I 3
L23 2 I 3
L3
−
6 I 3
L23 12 I 3 L33 6 I 3
−
L23
2 2 I 3 θ 3 L3 θ 6 I 3 w 4 − 2 L3 4 θ 4 I 3 L3 0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
76
School of Civil, Urban & Geosystem Eng., SNU 5.3 Frame Problems 5.3.1 M ember Stif fn ess M atri x
f y R ,δ R y
f y L ,δ L y m L ,θ L
m R ,θ R
f x L ,δ L x
f x R ,δ xR
Force-Displacement Relation at Member Ends
- Beam action m L
=
4 EI e
m R
=
2 EI e
L y
f
R y
f
=
Le Le
θ Le +
2 EI e
θ Le +
4 EI e
M Le + M R e Le
Le
Le
6 EI e
θ Re +
6 EI e
Le
=
M Le + M R e
=−
θ Re +
6 EI e 2 e
L
=−
2 e
L
2 e
L
θ Le +
6 EI e 2 e
L
δ L y −
6 EI e
δ L y −
6 EI e
6 EI e
L
L
L
6 EI e 2 e
L
δ R y
2 e
θ Re +
2 e
θ Le −
2 e
δ R y
12 EI e 3 e
L
θ Re −
δ L y −
12 EI e 3 e
L
12 EI e 3 e
L
δ L y +
δ R y
12 EI e 3 e
L
δ R y
- Truss action
=−
f x R
=
L e
V
EA
f x L
L EA L
L (δ R x − δ x )
(δ R x
− δ L x )
= V = 0 R e
Member Stiffness Matrix
f x L 0 Ae L 12 I e f y 0 L2e 6 I e m L 0 E Le = − A 0 f R Le e 12 I e 0 x − L2e R 6 I e 0 f y Le m R
0 6 I e Le 4 I e
−
0 6 I e Le
2 I e
− Ae
0 12 I e
0
−
0
−
Ae 0 0
L2e 6 I e
Le 0 12 I e L2e 6 I e
−
Le
δ L x 0 L 6 I e δ y Le θ L 2 I e e or 0 R 6 I e δ x − Le R 4 I e δ y θ R e
(f ) e = [k ]e ( ) e
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
77
School of Civil, Urban & Geosystem Eng., SNU
Transformation Matrix
V y
v y
v x
è ¨
V x
θ
è ¨ ç
= cos θv x − sin θv y V cos θ − sin θ x V y = sin θv x + cos θv y → V y = sin θ cos θ 0 M = m M 0 V x
0 v x
0 v y
1 m
(V ) = [ γ ]T ( v) → ( v ) = [γ ](V )
Member End Force e
F x1 cos θ − sin θ 1 F y sin θ cos θ 1 0 0 M 2 = 0 F x 0 F y2 0 0 M 2 0 0
0
0
0
0
0
0
1 0 0 0
0 f x L
e
L 0 f y
0 0 0 m L f R → cos θ − sin θ 0 x R sin θ cos θ 0 f y 0 0 1 m R
(F ) e
= [Γ ]T (f )e
Member End Displacement e
δ L x cos θ L δ y − sin θ θ L 0 Re = δ x 0 δ R y 0 θ R 0 e
sin θ
0
cos θ
0
0
1
0
0
0
0
0
0
0
0
0 ∆1 x
0
0
0
e
∆1 y 0 0 0 θ1 e e ∆2 → (δ) = [Γ](∆ ) cos θ sin θ 0 x 2 − sin θ cos θ 0 ∆ y 0 0 1 θ1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
78
School of Civil, Urban & Geosystem Eng., SNU
Member Stiffness Matrix in Global Coordinate
(F) e (F)
e
= [Γ]T (f )e = [Γ]T [k ]e (
)e
= [K ] ( ) where [K ] e
e
e
= [Γ]T [k ]e [Γ](
[K 11 ]e = e [K 21 ]
)e
[K 12 ]e
[K 22 ]e
Nodal Equilibrium & Compatibility Th e same as the truss pr oblems.
Global Stiffness Matrix
(P) = [C]T (F) = [C]T [K ](∆) = [C]T [K ][C](u) (P )= [K ](u)
where
[K ] = [C]T [K ][C]
Direct Stiffness Method Th e same as the truss pr oblems.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
79
School of Civil, Urban & Geosystem Eng., SNU
Chapter 6 Buckling of Structures
θ L
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
80
School of Civil, Urban & Geosystem Eng., SNU 6.0 Stability of Structures
Q
L
Stable state
K θ L × L > Qθ L → KL > Q The structure would return its original equilibrium position for a small perturbation in θ.
Critical state
K θ L × L = Qθ L → KL = Q
Unstable state
K θ L × L < Qθ L → KL< Q The structure would not return its original equilibrium position for a small perturbation in
θ.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
V+dV
dw dx
dx
Q
Equilibrium for vertical force
(V + dV ) − V + qdx = 0 →
dx 2
dx
dw
− P
2
dx 2
dM dx
−Q
dw dx
=
V
= −q
d 2 w dx 2
y +
du dx
Stress-strain relation (Hooke law)
d 2 w dx
2
y+ E
du dx
Definition of Moment
M =
d 2 w
∫ σ ydA = ∫ E ε ydA = −∫ ( E dx
A
=0→
Strain-displacement relation
σ = E ε = − E
dx
d 2 w
−Q
ε=−
=− q
Elimination of shear force
d 2 M
dx
Equilibrium for moment
( M + dM ) − M − Vdx + qdx
dV
A
2
y
2
− E
A
du dx
y ) dA = − EI
d 2 w dx 2
Beam Equation with Axial Force
EI
d 4 w dx 4
+Q
d 2 w dx 2
=
q
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
82
School of Civil, Urban & Geosystem Eng., SNU 6.2 Homogeneous Solutions
Characteristic Equation for P > 0
w = e λ x e λ x (λ4
+ β2λ2 ) = 0 → λ = ±βi,
EI
+ Cx + D
Exponential Function with Complex Variable
e e
0
Q
Homogeneous solution
w = Aeβix + Be −β ix
β2 =
where
ix
= 1 + ix +
− ix
i2
x 2!
= 1 − ix +
2
+
(−i ) 2 2! 1
i3
x 3!
x
2
i4
4
( −i ) 3
3
3
+
+
x 4!
3!
x
+ +
i5
x 5!
5
( −i ) 4 4!
+ x
i6
x 6 6!
4
+
+L
( −i ) 5 5!
x
5
+
( −i ) 6 6!
x 6
+L
1 1 x 2 + x 4 − x 6 + L) = 2 cos x 2! 4! 6! i i i = 2(ix − x3 + x5 − x 7 + L) = 2i sin x 3! 5! 7!
e ix
+ e− ix = 2(1 −
e ix
− e − ix
e ix
= cos x + i sin x
, e− ix
= cos x−
i sin x
Homogeneous solution
w = A(cos β x + i sin β x ) + B (cos β x − i sin β x ) + 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
Characteristic Equation for P < 0
w = eλ x e λ x (λ4
− β2λ2 ) = 0 → λ = ±β,
0
where
β2 =
Q EI
Homogeneous solution for P < 0
w = Aeβ x + Be −β x
= ( A + B)
eβ x
+ Cx + D
+ e −β x
eβ x
− e −β x
+ ( A − B) 2 2 = A cosh β x + B sinh β x + Cx + D
+ Cx + D
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
83
School of Civil, Urban & Geosystem Eng., SNU Simple Beam
Q
Q
− Boundary Condition w(0) = A + D = 0 , w′′(0) = − Aβ2
= 0 → A = 0 w′′( L ) = − Bβ 2 sin β L = 0 → B = C = 0
w( L) = B sin β L + CL = 0 ,
− Characteristic Equation A = B = C = D = 0 → w= 0 (trivial solution) or
β L = nπ → Q =
n 2 π 2 EI
L2 nπ w = B sin β x = B sin x L
, n = 1,2,3L
Fixed-Fixed Beam
Q
Q
− Boundary Condition w(0) = A + D
=0 w′(0) = β B + C = 0 w( L) = A cos β L + B sin β L + CL + D = 0 w′( L ) = − Aβ sin β L + Bβ cos β L + C = 0 − Characteristic Equation 0 1 0 β cos β L sin β L − β sin β L β cos β L
1
0 0 1 0 0 B 0 β Det ( = → cos β L 1 C 0 sin β L 0 D 0 − β sin β L β cos β L
0 1 L
1 A
1
0
0
1
0
β sin β L β cos β L
1
0
L
1
1
0
cos β L
− β sin β L
β = sin β L β cos β L
1
0
L
1
1
0
0
−
cos β L
− β sin β L
β sin β L β cos β L
0
1
1
0
L 1
) = 0 1 0
1 L 1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
84
School of Civil, Urban & Geosystem Geosys tem Eng., SNU
− β − (−β cos β L) − (−β(cos β L + β L sin β L) + β) = β(2 cos β L − 2 + β L sin β L) = 0 2 cos β L − 2 + β L sin β L = 2(cos β L − 1) + β L sin β L = β L β L β L − 4 sin 2 + 2β L sin cos =
2 β L β L β L sin ( cos 2 2 2
2
− sin
2
β L 2
) = 0 → sin
β L 2
=0
β L
or
2
cos
β L 2
− sin
β L 2
=0
− Eigenvalues Symmetric modes sin
β L 2
=0 →
w(0) = A + D
β L 2
= nπ → Q =
4n 2 π 2 EI L2
,
n = 1,2,3L
= 0 , w′(0) = w′( L) = β B + C = 0,
A + D = 0 → A = − D → w = A(cos
2nπ L
w( L ) = A + CL + D = 0
x− 1) for A ≠ 0
Anti-symmetric modes
β L 2
cos
β L 2
− sin
β L 2
=0→
β L 2
= tan
β L 2
→Q =
8.18π 2 EI L2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
85
School of Civil, Urban & Geosystem Geosys tem Eng., SNU
Cantilever Beam
Q
Q
− Boundary Condition w(0) = A + D
=0 w′(0) = β B + C = 0 M ( L ) = − EI w′′( L ) = − EI (− Aβ 2 cos β L − Bβ 2 sin β L ) = 0 V ( L ) = − EI
Q
=
n 2 π 2 EI 4 L2
d 3 w dx 3
dw
− P
dx
=0
n = 1,2,3L
,
6.3.Homogeneous and Particular solution w = wh EI
d 4 ( wh
+ w p )
dx 4
+Q
+ w p = A cos β x + B sin β x + Cx + D+
d 2 ( wh
+ w p )
dx 2
= EI = EI
d 4 wh dx 4 d 4 w p dx 4
+Q +Q
d 2 wh dx 2 d 2 w p dx 2
w p
+ EI
d 4 w p dx 4
+Q
d 2 w p dx 2
=q
Four Boundary Conditions for Simple Beams w(0) = A + D + w p (0) = 0 , M (0) = − EI w′′(0) = − EI (− Aβ 2
+ w p′′ (0)) = 0
w( L) = A cos β L + B sin β L + CL + D + w p ( L) = 0 M ( L) = − EI w′′( L ) = − EI (− Aβ 2 cos β L − Bβ 2 sin β L + w p′′ ( L)) = 0 1 0 − β2 0 cos β L sin β L 2 2 − β cos β L − β sin β L
0 0 L 0
w p (0) ′ ′ w ( 0 ) 0 B p + = 0 → KX + F = 0 1 C w p ( L ) w′′ ( L) 0 D p 1 A
The homogenous solution is for the boundary conditions, while the particular solution is for the equilibrium.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
86
School of Civil, Urban & Geosystem Geosys tem Eng., SNU 6.4.Energy Method
Total Potential Energy l
l
1 d 2 w 1 d 2 w dw dw EI 2 dx − Q dx− 2 2 0 dx 2 0 dx dx dx
∫
Π=
∫
l
∫ wqdx 0
∆
L − ∆
∫
L =
L − ∆
∫
ds =
0
L − ∆
∫
L = L − ∆ +
2
0
0
(1 +
0
( w′) dx → ∆ =
L − ∆
1
∫
2
2
0
+ w)
l
Π =
1 d 2 ( w e 2∫
dx
0
=
1 2
l
∫
(
d 2 w e
0
dx 2
l
− ∫ w qdx + 0
EI
2
EI 1 2
l
∫
(
0
d 4 w e
( w′) dx ≈ 2
1 2 1
( w′) 2 ) dx
L
( w′) dx for ∆ << L 2∫ 2
0
=Π +
1 2
l
∫ 0
(
−
d 2 w dx 2
EI
d 2 w
d 2 w
dx 2
dx
−
dx 2
d 2 w e dx 2
−
dx −
dw e
Q
dx
EI
l
1 d ( w e 2∫
d w dx
d w dx
l
Q
l
∫
∫
) dx − w qdx + ( e
0
Q
d w dx
− q)dx + Q
+ w)
dx
0
dw e
+Q
dx 4
0
e
dx 2
2
dx
dx 2 d 2 w
+ w)
d 2 ( w e
d 2 w e
l
= Π + ∫ w ( EI e
d w dx
0
d ( w e
d 2 w e dx 2
+ w)
dx EI
l
∫
dx − ( w e
+ w )qdx
0
d 2 w dx 2
−
dw e dx
Q
d w dx
) dx
) dx
1 2
l
∫ 0
(
d 2 w dx 2
EI
d 2 w dx 2
−
d w dx
Q
d w dx
) dx
)dx
The principle of minimum potential energy holds if and only if l
d 2 w
∫ ( dx
2
EI
0
∫
1 + ( w′) dx ≈
Principle of the Minimum Potential Energy h
1
L − ∆ 2
Q
d 2 w dx
2
−
d w dx
Q
d w dx
)dx
> 0 for all possible w
The principle of the minimum potential energy is not valid for the following cases. l
d 2 w
∫ ( dx 0
2
EI
d 2 w dx
2
−
d w dx
Q
d w dx
)dx ≤ 0 for some w
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
87
School of Civil, Urban & Geosystem Eng., SNU
The critical status of a structure is defined as l
∫
d 2 w
(
dx 2
0
EI
d 2 w dx 2
−
d w dx
Q
d w
) dx
dx
=0
Approximation n
− Approximation of displacement:
w
= ∑ ai g i i =1
− Critical Status l
n
n
0
i =1
j =1
( ∫ (∑ ai g i′′) EI ∑ n
n
n
0
i =1
j =1
a j g j′′ ) dx − P ∫ (∑ ai g i′ )∑ (
l
n
l
n
a j g j′ ) dx
l
n
n
=
n
n
n
′ g i′′dxa j − Q ∑∑ a∫i g i′ g j′ dxa j = ∑∑ ai K ij a j − Q ∑∑ ai K ijG a j = ∑∑ a∫i g i′ EI i =1 j =1
i =1 j =1
0
( a) T (K − K G )(a ) T
i =1 j =1
0
i =1 j =1
= 0 → Det (K − K G ) = 0
Example - Simple Beam
Q
w = ax( x − l ) → g 1′ = 2 x − l , g 1′′= 2
− with a parabola: l
l
′ g ′′dx = ∫ 2 EI 2dx = 4 EIl ∫ g ′ EI i
i
0
0
l
l
∫ g ′ EI g ′dx = ∫ (2 x − l ) i
l
2
i
0
dx
0
1
0
12 EI
Det ( 4 EIl − Q l ) = 0 → Qcr = 2 3 l 3
− with one sine curve:
w = a sin
π
l
∫
l
′ g i′′dx = EI ( ) g i′ EI l 0 π
l
4
∫
sin 2
l
∫ g ′ EI g ′dx = ( l ) ∫ cos i
i
0
0
π
π
π x l
2
(exact :
→ g 1′ =
3
3
π l
cos
π 2 EI l 2
EI
= 9.86
l 2
l 3
, error = 22%)
π x
π π x , g 1′′ = ( ) 2 sin l l l
π l dx = EI ( ) 4 l l 2
π x
π l dx = ( ) 2 l l 2
l Det ( EI ( ) − Q( ) ) = 0 → Q = l 2 l 2 4
l
2
1
π x
0
2
4
= ∫ (4 x 2 − 4 xl + l 2 )dx = ( − 2 + 1)l 3 =
π 2 EI l 2
(exact )
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
88
School of Civil, Urban & Geosystem Eng., SNU
Example – Cantilever Beam
Q
Q
− with one unknown: w = ax 2
→ g 1′ = 2 x , g 1′′ = 2
l
l
= ∫ 2 EI 2dx = 4 EIl ,
∫
′ g i′′dx g i′ EI
0
0
Det (4 EIl − Q
4 3
l ) = 0 → Qcr
=
3
l
∫
g i′ EI g i′dx
0
l
= ∫ 4 x 2 dx = 0
3 EI
(exact :
l 2
π 2 EI 4l 2
4 3
l 3 EI
= 2.46
l 2
, error = 22%)
− with two unknowns: w = ax 2
+ bx3 → g 1′ = 2 x , g 2′ = 3 x 2 , g 1′′ = 2 , g 2′′ = 6 x
l
G 11
K
= ∫ 4 x
2
dx =
0
4 3
l
3
G 12
l , K
= K = ∫ 6 x G 21
= ∫ 4dx = 4l , K 12 = K 21 = ∫12 xdx = 6l
2
4l Det ( EI 2 6l
α=
4
l
4
G 22
l , K
3 1 40l − Q 30 4 12l 3 45l
6l 2
26 2 l 12 8
45l
− 180l 12
45l 4
= ∫ 9 x 4dx = 0
9 5
l 5
l
, K 22
0
(4l − 40l 3 α )(12l 3
±
6
l
0
26l 6
dx =
0
l
K 11
3
= ∫ 36 x 2dx = 12l 3 0
4l − 40l 3 α
− 45l 4 α 1 Q ) 0 0 , = → = α = 30 EI 54l 5 6l 2 − 45l 4 α 12l 3 − 54l 5 α 6l 2
− 54l 5 α) − (6l 2 − 45l 4 α) 2 = 0 → 4l 4 − 52l 6 α + 45l 8 α 2 = 0 =
26 ± 22.27 2
45l
=
0.0829 2
l
or
1.0727 2
l
→ Qcr =
EI EI 2.487 2 or 32.181 2 l l
EI EI Qexact = 2.49 2 (error = 1.2%) or Qexact = 22.19 2 (error = 45%) l l
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
89
School of Civil, Urban & Geosystem Eng., SNU 6.5 Approximation with the Homogeneous Beam Solutions
Homogeneous Solution of Beam u2i − 1
u2i + 1 P
u2i + 2
u 2i
i-th member
δ L y θ L z L L R R = 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 ) δ y R θ z
wi
3 x 2
=1−
N 1
L2
2 x 3
+
, N 2
L3
2 x 2
= x −
L
+
x 3
, N 3
L2
3 x 2
=
L
−
2 x 3 L2
, N 4
=−
x 2 L
+
x 3 L2
6.5.1 Beam Analysis
Total Potential Energy p
Π=∑ i =1 p
=∑ i =1
=∑ i =1
=
∫
2
dx 2
0 l
1
EI
d 2 wi
( 2∫ dx
2
d 2 wi dx 2
T
) EI
1 2
l
(∆ i )
d 2 N
∫ ( dx
T
2
dx −
dx
2
T
) EI
0
p
∑ (∆ ) 2 i
T
p
∑ 2 Q∫
dx −
= = =
2 1 2 1 2
K i0
(u)
0
1
d 2 N dx
2
G i
dwi dwi dx dx l
dwi
dx( ∆ i ) −
p
p
l
∑ ∫ w q dx i
i
i =1 0
T
dw i dx
0
dx −
l
∑ ∫ (w ) i
T
d N
i
i =1
T
qi dx
i =1 0
l
1
p
∑ 2 Q(∆ ) ∫ ( dx ) 0
T
d N dx
dx (∆ i ) −
l
p
∑ ( ∆ ) ∫ (N ) T
T
i
i =1
0
p
∑ (∆ )
T
i
(f i )
i =1
∑ [C ]
T
i
([K ] − [K ])[C i ](u) − 0 i
G i
i =1
p
∑ (u)
T
[C i ]T (f i )
i =1
p
(u) T
dx −
∑ 2 Q∫ ( dx )
([K ] − [K )(∆ i ) − 0 i
p
T
p
l
i =1
i =1
1
1
i =1
d 2 wi
0
p
1
l
2 1 d wi
p
∑ [C ] [K ][C ](u) − (u) ∑[C ] T
i
T
i
i
i =1
i
T
(f i )
i =1
(u) T [K ](u) − (u) T ( P)
=
EI i Li
6 12 6 12 − L2 Li Li L2i i 6 6 4 − 2 Li Li , 12 6 12 6 − 2 − − 2 L Li L L i i i 6 6 2 4 − L Li i
K Gi
1 6 1 6 − 5 L 10 5 Li 10 i 2 L L 1 1 i − i 10 30 = Q 10 15 6 1 6 1 − − 5 L 10 5 L 10 i i 1 − Li − 1 2 Li 10 30 10 15
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
qi dx
90
School of Civil, Urban & Geosystem Eng., SNU
Principle of Minimum Potential Energy for Q < Qcr
Π=
1 2
(u)T [K ](u) − (u)T ( P) =
1 2
n
n
n
∑ ∑ ui
i =1
K ij u j
j =1
− ∑ ui P i i =1
n ∂u j n 1 n ∂Π 1 n ∂u i n K ij u j + ∑ u i ∑ K ij = − ∑ u i P i ∑ u 2 u ∂u k 2 ∑ ∂ ∂ i =1 i =1 j =1 i =1 k j =1 k
= =
1
n
∑
K kj u j
2 j =1 1
n
∑
K kj u j 2 j =1
+ +
1 2 1
n
∑
u i K ik − P k
=
i =1 n
∑
K kj u j − P k 2 j =1
1
n
∑
K kj u j
2 j =1
+
n
1 2
n
∑ K u ki
i
− P k
i =1
= ∑ K kj u j − P k = 0
for k = 1, L , n → [K ](u)= ( P)
j =1
Calculation of the Critical Load
Det ([K ]) = Det ([K 0 ] − [K G ]) = 0
Frame Members
f x L 0 0 Ae L f y 12 I e 6 I e 0 L2e Le 6 I e 0 m L 4 I e E Le = ( A 0 0 f R Le e 0 − 12 I e − 6 I e x L2e Le R 6 I e 0 2 I e f y L e m R
Ae 0 0 Ae 0 0
−
0 12 I e
−
L2e 6 I e
Le 0 12 I e 2 e
L 6 I e
−
Le
0 0 0 6 1 0 Le 5 Le 10 1 2 Le 2 I e 0 10 15 − P 0 0 0 0 6 1 6 I − e 0 − 5 L 10 Le e 1 L 0 − e 4 I e 10 30 0 6 I e
P = EA
0 0 0 0 0 0
0 6
−
5 Le 1 10 0 6
5 Le 1
−
10
δ L x 0 δ L 1 y 10 Le θ L e − 30 ) 0 1 R − δ x 10 2 Le R δ 15 y θ R e
δ L x −δ R x Le
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
91
School of Civil, Urban & Geosystem Eng., SNU 6.6 Nonlinear Analysis of Truss f y R f y L
f x L
δ L y
δ L x
f x R
EA
f x L
=−
f x R
=
f
e
(f ) e
L
− δ L x )
(δ R x
− δ L x )
= − f = L y
0
−1
0
0
0
1
0
0
δ R y − δ L y l
f x R
0
0 0 R L 0 δ − δ + x x 0 1 0 0 l 0 0 0 − 1
= ([k ]e0 + f x R [k ] g e )(
)e
0 0 0 0
= ([k ]e0 + p e [k ] g e )(
0 δ x L
e
− 1 δ L y 0 δ R x R 1 δ y )e
Equilibrium Analysis
(f )ie (F ) e
L EA
(δ R x
Member Stiffness Matrix
f x L 1 L f y EA 0 f R = l − 1 x f y R 0
δ R x
Force – Displacement relation at Member ends
R y
δ R y
= [Γ]T (f )e = [Γ]T [k ]e (
= ([k ]e0 + )e
e pie [k ] g )( )ie
= [Γ]T ([k ]e0 + pie [k ] g e )[Γ](
) e = [K ]e ( ) e
Successive substitution
(F ) ie−1
≈ [Γ]T ([k ]e0 + pie−1[k ] g e )[Γ]( (P) = ([K ]0
) e = [K ]ie−1 ( ) e
+ [K ( pie−1 )]G )(u)i
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
