Structural Analysis I - SNUstrana.snu.ac.kr/lecture/struct1_2015/Notes/15_Note_All.pdf · Dept. of...
Transcript of Structural Analysis I - SNUstrana.snu.ac.kr/lecture/struct1_2015/Notes/15_Note_All.pdf · Dept. of...
Dept. of Civil and Environmental Eng., SNU
Structural Analysis I
Spring Semester, 2015
Hae Sung Lee
Dept. of Civil and Environmental Engineering Seoul National University
yδ yf
zδ zf
xδ xf
yM yθ
zM zθ xM xθ
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU
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Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU
Contents
1. Introduction
2. Reactions & Internal Forces by Free Body Diagrams
3. Principle of Virtual Work
4. Analysis of Statically Indeterminate Beams
5. Analysis of Statically Indeterminate Trusses
6. Analysis of Statically Indeterminate Frames
7. Influence Lines for Determinate Structures
8. Influence Lines for Indeterminate Structures
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU
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Dept. of Civil and Environmental Eng., SNU 1
Chapter 1
Introduction
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1.1 Mechanics of Material - Structural Mechanics
Problem
Calculate the reaction force at each support and draw the moment and shear force diagram for
the two-span beam shown in the figure.
Solution
Equilibrium Equation
qLRRRF cbay 20 =++→=∑
qLRRLRLRLqLM cbcba 220220 =+→=×−×−×→=∑
0022
0 =−→=×−×+×+×−→=∑ cacab RRLRLRLqLLqLM
qLRRLRLRLqLM babac 220220 =+→=×+×+×−→=∑
Since there are three unknowns in two independent equations, we cannot determine a unique
solution for the given structure, and thus we need one more equation to solve this problem.
The main issue of this class is how to build additional equations to analyze statically
indeterminate structures.
EI EI
q
Ra Rb Rc
L L
q
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1.2 Mechanics of Material
Governing Equation
– Left span
112
13
1
4
1''''
1 24dxcxbxa
EIqxwqEIw ++++=→=
– Right Span
222
23
2
4
2''''
2 24dxcxbxa
EIqxwqEIw ++++=→=
Boundary Conditions
– Left support
0)0()0( , 0)0( 111 =′′−== wEIMw
– Center support
)()( , )()( , 0)()( 212121 LwLwLwLwLwLw ′′=′′′−=′==
– Right support
0)0()0( , 0)0( 222 =′′−== wEIMw
Since there are eight unknowns with eight conditions, we can solve this problem.
Determination of Integration Constant
– Left Support
xcxaEI
qxwbwdw 13
1
4
11111 2402)0( , 0)0( ++=→==′′==
– Right Support
xcxaEI
qxwbwdw 23
2
4
22222 2402)0( , 0)0( ++=→==′′==
x x
q
w1 w2
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– Center Support
==
−==→
+=+
−−−=++
=++=
=++=
EIqLcc
EIqLaa
LaEI
qLLaEI
qL
cLaEI
qLcLaEI
qL
LcLaEI
qLLw
LcLaEI
qLLw
48
483
62
62
36
36
024
)(
024
)(
3
21
21
2
2
1
2
22
2
3
12
1
3
23
2
4
2
13
1
4
1
)32(48
33421 xLLxx
EIqww +−=≡
83 ,
83
2 112
11qLqxwEIVxqLxqwEIM +−=′′′−=+−=′′−=
Moment Diagram Shear Diagram Reactions
0.375qL
L83
+ -
+
0.625qL
-
0.375qL 0.375qL 1.25qL
0.125qL2
0.070qL2
L83
+ -
+
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1.3 Mechanics of Material + α
1.3.1 Main idea
Original Problem
Case I (Removal of the center support)
Case II (Application of the reaction force)
Original Problem = Case I + Case II
δ0+ δR=0
(compatibility condition) 1.3.2 Calculation of δ0
Governing Equation
dcxbxaxEI
qxwqEIw ++++=→= 234
0''''
0 24
q
q
δ0
δR
Rb
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Boundary (support) Conditions
– Left Support
0)0()0( , 0)0( 000 =′′−== wEIMw
– Right Support
0)2()2( , 0)2( 000 =′′−== LwEILMLw
Determination of Integration Constant
– Left Support
00)0( , 0 0)0( 00 =→=′′=→= bwdw
– Right Support
+=
−=→
=+
=++→
=′′=
EILqc
EILqa
LaEILq
LcLaEILq
LwLw
24)2(
12)2(
0)2(62
)2(
0)2()2(24
)2(
0)2( 0)2(
32
34
0
0
))2()2(2(24
3340 LxLxx
EIqw +−=
EILqLLLLL
EIqLw
384)2(5))2()2(2(
24)(
4334
00 =+−==δ
1.3.3 Calculation of δR
Governing Equation
dcxbxaxwEIw RR +++=→= 23'''' 0
Boundary (support) Conditions
– Left Support
0)0()0( , 0)0( =′′−== RRR wEIMw – Mid-span
2
)()( , 0)()( bRRR
RLwEILVLwL −=′′′−==′=θ
Determination of Integration Constant
– Left Support
00)0( , 0 0)0( =→=′′=→= bwdw RR Structural Analysis Lab.
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Dept. of Civil and Environmental Eng., SNU 7
– Mid-span
)3(12
123
12
26
03
2)(
0)( 23
2
2
xLxEI
Rw
EILR
c
EIR
aR
aEI
caLR
LwEI
Lwb
Rb
b
bbR
R
−=→
−=
=→
=
=+→
=′′′
=′
EIRL
EIRL
Lw bbRR 48
)2(122
)(33
−=−==δ
1.3.4 Final Solution
Reaction at Supports
qLRRR cba 2=++
qLRR cb 22 =+
δ0+ δR=0 → 048
)2(384
)2(5 34
=−EI
RLEILq b → qLRb 8
10=
qLRR ca 83
==
Moment
xqLxqxqLxLxqwEIwEIMMM RR 83
285)2(
222
00 +−=−+−=′′−′′−=+=
Shear
8
38
5)(00qLqxqLLxqwEIwEIVVV RR +−=−+−=′′′−′′′−=+=
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1.4 Structural Mechanics
Original Problem
Case I (Removal of the center support)
Case II (Application of the reaction force)
Original Problem = Case I + Case II
δ0+ δR=0
Principle of Virtual Work
EILqdx
EIMML
R
384)2(5 42
0
00 −== ∫δ ,
EILRdx
EIMM b
LRR
R 48)2( 32
0
== ∫δ
Solution
δ0+ δR=0 → 048
)2(384
)2(5 34
=+−EI
RLEILq b → qLRb 8
10=
RbL/2
Rb
q
q
qL2/2
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– Moment
– Shear
+
-
+
-
+
+ -
+ -
=
0.070qL2
5qL2/8
Rb
+
=
0.125qL2 L83
+ -
+
qL2/2
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1.5 지 점 (Supports)
고정단 (fixed support)
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회전단 (hinge support)
이동단 (roller support)
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1.6 구조물의 2차원 이상화
주 구조물 (Main Structure)
가로 보 (Cross Beam)
세로 보 (Stringer )
Cross Bracing (Wind Bracing)
지 점 (Support)
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Truss
Beam
절점(Joint)
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Frame
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1.7 Force and Displacement
Real 3-D Structures
– 3 force components and 3 moment components
– 3 displacement components and 3 rotational components
Beam Idealization
– Vertical force and Moment on z-axis
– Vertical displacement and rotational angle w.r.t. z-axis
Plane Truss Idealization
– Vertical and horizontal force
– Vertical and horizontal displacement
x
y
z
yδ yf
zδ zf
xδ xf
yM yθ
zM zθ xM xθ
22 , wV
22 , θM
11 , wV
11 , θM
33 , wV
33 , θM
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Plane Frame Idealization
– Vertical, horizontal force and moment w.r.t. z-axis
– Vertical, horizontal displacement rotational angle w.r.t. z-axis
xδ xf
yδ yf
xδ xf
yδ yf zM
zθ
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1.8구조물의 안정 (Stability of Structures)
내적 안정 (Internal Stability)
어느 한계 내의 크기의 어떠한 하중의 작용을 받더라도 형상이 허물어 지지
않는 구조물의 상태
외적 안정 (External Stability)
어느 한계 내의 크기의 어떠한 하중의 작용을 받더라도 구조물이 움직이지 않
는 상태
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Chapter 2
Reactions & Internal Forces
by Free Body Diagrams
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2.1 Free Body Diagram
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It is impossible to draw too many free-body diagrams.
Time spent in doing so is never wasted
- C. H. Norris & J. B. Wilbur & S. Utku -
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2.2 Reactions
Beams
0 0 =−+→=∑ PRRF BAV
00 =−→=∑ LRPaM BA (Clockwise +)
PLaRB = , P
LbRA =
PRRF BAV =+→=∑ 0
0)(0 =−+→=∑ LRaLPM BA (Clockwise +)
PLaRB )1( += , P
LaRA −=
L
P
a b
RA RB
L
P
a
RA RBaL
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Truss
PHPHF AAH −=→=+→=∑ 00 ,
00 =−+→=∑ PRRF BAV
0)3(0 =−+→=∑ aRPaPaM BA (Clockwise +)
PRA 31
= , PRB 32
=
Frame
0
02
=+=
=−+=
∑∑
BAH
BAV
HHF
qLRRF
∑ =−−= 02
LHLRM BBRh
∑ =−−= 0422LqLLHLRM AA
Lh BB HR 2−= ,
42 qLHR AA +=
822
42 qLHHqLHqLH BABA =−→=−+
16
16qLH
qLH
B
A
−=
= →
8
83
qLR
qLR
B
A
=
=
P RA RB
P
HA
L
L
HA HB
RA RB
q
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2.3 Internal Forces in Framed Structures
Axial Force
Shear Force
Bending Moment
Torsion
+
+
+
+
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2.4들보 (Beam)
Reactions
q
RA=qL/2 Rb= qL/2 Structural Analysis Lab.
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Dept. of Civil and Environmental Eng., SNU 26
Free Body Diagram for Shear and Moment
qxqLqxRVVqxRF AxxAV −=−=→=−−=∑ 20
22
02
2qxxqLMMxqxxRM xxAx −=→=−−=∑
Shear Force and Moment Diagrams
q
RA=qL/2 RB= qL/2
RA RB
RA
x
RB
Mx
Vx
+
qL/2 qL2/8
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Deflected Shape
2.5 Gerber Systems
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2.5.1 Internal Forces in a Gerber Beam - I
Free Body Diagram
PRLPLRM HHC 320
2430 =→=×−×→=∑
PRPRRF CCHv 3100 =→=−+→=∑
PRLRLRM BHBA 650
450 =→=×+×−→=∑
PRPRRF ABAv 610
320 −=→=−+→=∑
L/4 P
RH
RA RB
RC
P
P/6 5P/6 P/3
P
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Shear Force i) Lx ≤≤0
ii) LxL23
≤≤
iii) LxL 223
≤≤
Bending Moment i) Lx ≤≤0
P/6
V= -P/6
P/6 5P/6
V= 2P/3
P/6 5P/6
P
V= -P/3
+
- - 2P/3
P/3 P/6
P/6 P/6
Mx= Px/6
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ii) ,
iii)
Deflected Shape
P/6 5P/6
2P/3
Mx=
P/6 5P/6
P
P/3 P/3
Mx=
+ -
PL/6
PL/6
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2.5.2 Internal Forces in a Gerber Beam - II Free Body diagram Shear
Moment
q
2ql
2ql
2ql
2ql
2
2ql
2
2ql
2ql
2ql
+
L
q
L L
8
2ql
2
2ql 2
2ql
+
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2.6 트러스 (Truss)
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Assumption
1. All joints are hinges.
2. All members are straight.
3. Small deformatiom
4. The external loads are applied only at joints.
Characteristics of truss
– By the 2nd , 3rd and 4th assumptions
– By the 1st assumption
– No bending moment and shear force are induced in all members in a truss structure.
– Only axial forces are the internal forces in a truss.
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2.6.1 Internal Forces in Howe Truss
At U1 and U3
At L1
2 ,
22
022
2
022
23
3
23 PFPFPF
FF=−=→
=+
=+
At L2
2 , 0- , 0 265625
PFFPFFFPF ===→=+=−
F1 F3
F2
P/2
F5
F2 F6
P
F4=0
F1=0
F8 =0
F9=0
U2
H 1
2
3
4
5
6
7
9
L1 L2
L3
U1 U3
P P/2 P/2
8
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At U2
=+−
−==→
=−−−
=++−
021
21
22
022
22
022
22
37
753
8734
PPP
PFF
FFF
FFFF
At L3
Axial Force Diagram
2P
2
P
F5
F4 F8
F3 F7
2P
0
P/2
P/2
P/2 P/2
P 0
0 0
0
P P/2 P/2
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Equilvalent Beam Action
Deflected Shape
2.6.2 Internal Forces in Warren Truss
V=P/2
P/2
1 P
2 3
4
5
6
7
8
9
10
11
L1 L2 L3
U1 U2 U3
L4
x
PL/4 M=Px/2
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At L1
PFPFPF
FF
32 ,
322
03
222
022
12
2
12
=−=→
=+
=+
At U1
PFPFFFF
FFF
34,2
32
022
22
022
22
423
32
432
−==−=→
=+
=++−
At L2
PFPFFFFF
PFF==→
=+−+
=−+65
5361
35
,32
022
22-
022
22
F2
F1
2P/3
F4
F2 F3
F3
F1 F6
P
F5
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At U2
PFPFFF
FFFF
32,
32
022
22
022
22
87
75
7584
−=−=→
=−−
=+−+−
At L3
3,
32
022
22
022
22
109
97610
97 PFPFFFFF
FF==→
=+−−
=+
At U3
PFFF
FFF
32
022
22
022
22
11
119
8119
−=→
=−−
=−+−
F8
F5 F7
F4
F7
F6 F10
F9
F11
F8
F9
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At L1
OK
Deflected Shape Equivalent Moment
1 2 3 4 5 6
1.
2.
3.
4.
5.
6.
P/3
P/3
P
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2.6.3 Method of Sections
PFLFLPM L 340
232
442 −=→=×+×=∑
PFLFLPLPMU =→=×−×−×=∑ 662 0222
332
PFFPPFV 320
22
32
55 =→=×+−=∑
P
3P
P32
1
2 3
4
5
6
7
8
9
10
11
L1
L2 L3
U1
U2 U3
L4
Cut out
P P
32
F5
F4
F6
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2.7 프레임 (Frame)
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2.7.1 Internal Forces in a Frame
Reactions
Freebody Diagram
H
qH
L
LqH2
2
q
LqH2
2
2
2qH
qH
LqH2
2
LqH2
2
2
2qH
LqH2
2
LqH2
2
LqH2
2
LqH2
2
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Axial, Shear and Moment diagram
Deflected Shape
+ - Axial
-
+ Shear
+
Moment
+
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Dept. of Civil and Environmental Eng., SNU 45
2.7.2 Internal Forces in a 3-hinged Frame
Reactions (+:Clockwise for mement)
02
=+
=+
BA
BA
HH
qLRR
∑ = 0RhM : BBBB HRLHLR 20
2−=→=−−
∑ = 0LhM :
420
422qLHRLqLLHLR AAAA +=→=−−
082
24
2
=+
=−→=−+
BA
BABA
HH
qLHHqLHqLH
16
16qLH
qLH
B
A
−=
= →
8
83
qLR
qLR
B
A
=
=
L
L
HA HB
RA RB
h
q
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
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Freebody Diagram
Axial, Shear and Moment diagram
16qL
16qL
8qL
16qL
83qL
16qL
8qL
83qL
8qL
16qL
16qL
16
2qL 16
2qL
16
2qL 16
2qL
16qL 16
qL
- +
- 8
qL
+
83qL
Shear
83qL
8qL
16qL
- -
-
Axial
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Deflected Shape
- -
-
Moment
- -
M
V
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2.8 Arches
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2.8.1 Three Hinged Arch
Arch Curve : )( 22
2 xllhy −=
Reactions
2
2PR
PR
B
A
=
= ,
hPlH
hPlH
B
A
2
2
=
=
Freebody Diagram
hPlH
PV
2
2
=
=
0)( =++−− xLRyHM AA
)(2
)(2
)(2
)(22
2
222
xlxl
P
xlPxllh
hPl
xlPyh
PlM
+=
++−−=
++−=
8maxPlM −=
HA
RA
HB
RB
P
h
2l
x
y
HA
RA
V
H M
y
l+x
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 50
Axial force and Shear Force
Deflected Shape
V
H
A S θ
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 51
2.8.2 Zero Moment Arch I
Reactions
2
2PR
PR
B
A
=
= ,
hPlH
hPlH
B
A
2
2
=
=
Freebody Diagram
hPlH
PV
2
2
=
=
0)( =++−− xLRyHM AA
)(
0)(22
xllhy
xlPyh
PlM
+=
=++−=
HA
RA
HB
RB
P
h
2l
x
y
H
HA
RA
V M
y
l+x
x y
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 52
2.8.3 Zero Moment Arch II
Reactions
2
2qlR
qlR
B
A
=
= ,
hqlH
hqlH
B
A
2
22
2
=
=
Freebody Diagram
02
)()()( =+
+−++−−xlxlqxlRyHM AA
)(
02
)()()(2
222
2
xllhy
xlxlqxlqlyh
qlM
−=
=+
+−++−=
HA
RA
HB
RB
q
h
2l
x
y
HA
RA
V
H
M
y
l+x
q
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 53
Chapter 3
Principle of Virtual Work
The principle of virtual work is the most important subject in the area of the structural analysis !!!!
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 54
3.1 Beam Problems
3.1.1 Governing Equations
Equilibrium for vertical force
qdxdVqdxVdVV −=→=+−+ 0)(
Equilibrium for moment
Vdx
dMdxqdxVdxMdMM =→=+−−+ 02
)(
Elimination of shear force
qdx
Md−=2
2
Strain-displacement relation
ydx
wd2
2
−=ε
Stress-strain relation (Hooke law)
ydx
wdEE 2
2
−=ε=σ
Definition of Moment
2
22
2
2
dxwdEIdAy
dxwdEydAEydAM
AAA
−=−=ε=σ= ∫∫∫
Beam Equation
qdx
wdEI =4
4
M M+dM
V V+dV
q
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 55
Modelling of Concentrate loads - Dirac delta functions
0lim→ε
= = )( ξ−δ x
1221lim)0
210(lim)(
00
00
=εε
=+ε
+=ξ−δ→ε
ε+ξ
ε+ξ
ε−ξ
ε−ξ
→ε ∫∫∫∫ll
dxdxdxdxx
)()(2
)()(lim)(21lim
)0)(21)(0)((lim)()(
00
00
0
ξ=ξ′=ε
ε−ξ−ε+ξ=
ε=
+ε
+=ξ−δ
→ε
ε+ξ
ε−ξ→ε
ε+ξ
ε+ξ
ε−ξ
ε−ξ
→ε
∫
∫∫∫∫
fFFFdxxf
dxxfdxxfdxxfdxxxfll
3.1.2 Principle of Virtual Work (Beam)
Equilibrium equation in an integral form
0)(0
2
2
=+∫ dxqdx
Mdwl
Integration by part twice
dxqwdxMdx
wdMdxwd
dxdMw
dxqwdxdx
dMdxwd
dxdMw
llll
lll
∫∫
∫∫
−=+−
−=−
002
2
00
000
llll
MVwdxqwdxMdx
wd00
002
2
θ+−−= ∫∫
In case w is a displacement field of the same structure caused by another load case q ,
then the boundary terms vanish since either displacement or reaction should be zero at a
boundary (support).
ξ
∞
ξ
ε21
2ε
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 56
Principle of virtual work
dxEIMMdxM
dxwd ll
∫∫ −=00
2
2
dxqwdxEIMM ll
∫∫ =00
→ extWW δ=δ int
Equilibrium equation for load case q
0)(0
2
2
=+∫ dxqdx
Mdwl
Virtual work expression
dxqwdxEIMM ll
∫∫ =00
Betti-Maxwell’s Reciprocal Theorem
dxEIMMdx
EIMM ll
∫∫ =00
→ dxqwdxqwll
∫∫ =00
Calculation of displacement for the load case q
dxqwdxEIMM ll
∫∫ =00
In case q system represents a single unit concentrated load applied at the position where
you want to calculate the displacement for q system.
)()( 00
00
xwdxxxwdxEIMM ll
=−δ= ∫∫ → dxEIMMxw
l
∫=0
0 )(
3.1.3 Example
A simple beam subject to an uniform load
– Moment of load case q
q
ql2/8
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 57
– Moment of load case
– Deflection at the center of the span
or from the integration table,
Values of Product Integrals
l/4
1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 58
3.1.4 Conservation of Energy
Equilibrium and Conservation of Energy
– Equilibrium Equation
qdx
wdEI =4
4
– External work
int
ll
lll
lll
ll
ext
WdxEIMdx
dxwdEI
dxwd
MwVdxdx
wdEIdx
wd
dxwdEI
dxdw
dxwdwEIdx
dxwdEI
dxwd
dxdx
wdwEIwqdxW
===
θ+−=
−+=
==
∫∫
∫
∫
∫∫
0
2
02
2
2
2
000
2
2
2
2
02
2
03
3
02
2
2
2
04
4
0
21
21
][21
][21
21
21
Conservation of Energy in each load case
dxwqdxEIM ll
∫∫ =00
2
21
21 , dxqwdx
EIM ll
∫∫ =00
2
21
21
Two load cases are applied simultaneously.
dxqwqwdxEIMM
dxwqqwqwqwdxEIMdx
EIMMdx
EIM
dxqqwwdxEI
MM
ll
llll
ll
∫∫
∫∫∫∫
∫∫
+=
+++=++
++=+
00
00
2
00
2
00
2
)(21
)((21
21
21
))((21)(
21
w
q
w
q
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 59
External work for sequential loading (q first)
dxwqdxwqdxqwqwdxwq
dxwqqwqwqwdxqwqdxwwqdx
dxqqwwdxqwdxwqwqdx
llll
lll l
lll l
∫∫∫∫
∫∫∫ ∫
∫∫∫ ∫
=→+=
+++=++
++=++
0000
000 0
000 0
)(21
)((21
21
21
))((21
21
21
Principle of Virtual work
dxqwdxqwdxqwqwdxEIMM llll
∫∫∫∫ ==+=0000
)(21
3.1.5 General Conservation and Equilibrium
Conservation in General
∫ ∫ =+⋅−S V
fdVdS 0nv
– By divergence theorem,
∫∫ ⋅∇−=⋅−VS
dVdS vnv where ),,(),,(321 xxxzyx ∂
∂∂∂
∂∂
=∂∂
∂∂
∂∂
=∇ .
w
q
w
q
ww +
v n
dS
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 60
∫∫∫ ∫∫ =+⋅−∇=+⋅∇−=+⋅−VVS VV
dVffdVdVfdVdS 0)( vvnv
– Since the integral equation should hold for all systems,
0=+⋅∇− fv
– In a potential problem, the vector field of a system is defined by a gradient of a scalar
function referred to as a potential function
Φ∇−= kv
– The famous Laplace equation for a conservative system.
02 =+Φ∇=+Φ∇⋅∇=+⋅∇− fffv or
02
2
2
2
2
2
=+∂
Φ∂+
∂Φ∂
+∂
Φ∂ fzyx
Equilibrium in General
– Force Equilibrium: ∑∑∑ === 0zyx FFF or 0=∑F
∫ ∫ =+S V
dVdS 0bT or ∫ ∫ =+S V
ii dVbdST 0 for i = 1,2,3
Suppose nT ⋅= σ or ∑=
⋅==3
1jijiji nT nσσ
=
σσσσσσσσσ
=
3
2
1
333231
232221
131211
σσσ
σ ,
=
3
2
1
nnn
n
Divergence Theorem
0)( =+⋅∇=
+⋅∇=+⋅=+
∫
∫ ∫∫ ∫∫ ∫
Vii
V Vii
S Vii
S Vii
dVb
dVbdVdVbdSdVbdST
σ
σσ n for i = 1,2,3
Since the integral equation should hold for all systems in equilibrium,
03
1
321 =+∂
σ∂=+
∂σ∂
+∂σ∂
+∂σ∂
=+⋅∇ ∑=
ij j
iji
iiiii b
xb
zyxbσ for i = 1,2,3 or
0
0
0
3333231
2232221
1131211
=+∂σ∂
+∂σ∂
+∂σ∂
=+∂σ∂
+∂σ∂
+∂σ∂
=+∂σ∂
+∂σ∂
+∂σ∂
bzyx
bzyx
bzyx
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 61
– Moment Equilibrium: 0=∑ iM for i=1, 2, 3 or 0=∑M
0=+×+× ∫∫∫VVS
dVdVdS mfxvx
211231133223 , , σ=σσ=σσ=σ
– What is σ, and how is σ related to a potential function? : out of scope of this class !
3.1.6 Displacement on boundaries
lll
llll
MVwdxqwMVwdxqwdxEIMM
000
0000
θ−+=θ−+= ∫∫∫
)0()0()()()0()0()()(00
0
MlMlVwlVlwMVwdxEIMM ll
l
θ+θ−−=θ−=∫
By coinciding the positive direction of forces and displacement
dxEIMMMlMlVwlVlw
l
∫=θ+θ++0
)0()0()()()0()0()()(
Deflection of a cantilever beam subject to an end load
0)0( , 0)0( , 0)( , 1)( =θ=== wlMlV
q (real) system q (virtual) system
EIPlPll
EIlMM
EIldx
EIMMlw
l
3))((
33)(
3
310
=−−=== ∫
Or, you can obtain the same answer by assuming the unit concentrate load is applied at
just left side of the boundary.
Rotation of a cantilever beam subject to an end load
0)0( , 0)0( , 1)( , 0)( =θ=== wlMlV
q (real) system q (virtual) system
P
-Pl
1
-l
P
-Pl
1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 62
EIPlPl
EIlMM
EIldx
EIMMl
l
2)(1
22)(
2
310
−=−××===θ ∫
Rotation in the a body (or a structure)
– Modeling of a unit moment applied at x0
)()]2
()2
([1lim
))]2
((1))2
((1[lim
0000
0000
0
0
xdxdwxwxw
dxxxxxwdxEIMM
xx
ll
θ−=−=ε
+−ε
−ε
=
ε+−δ
ε−
ε−−δ
ε=
=→ε
→ε ∫∫
– by coinciding the positive direction of the rotational angle with that of the applied moment.
dxEIMMx
l
∫=θ0
0 )(
ε 1/ε 1/ε
x0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 63
3.2 Principle of Virtual Work in General
3.2.1 3-Dimensional Elastic Body
Rigid Body
If a real q-force system is acting on a rigid body is in equilibrium and remains in
equilibrium as the body is given any small displacement , the virtual work done by
the q-force system is equal to zero.
0)(∫ ∫∫ ∫ =+⋅=⋅+⋅=δS VS V
ext dVdSdVdSW fvwfwvw
Deformable Body
If a deformable body is in equilibrium under a real q force system while it is subjected to
small and compatible displacement caused by a virtual q force system, the external
virtual work done by the real q force system is equal to internal virtual work done by
the internal q stress !!!
∫ ∫ ⋅+⋅=δS V
ext dVdSW fwvw
∑∑∫∑∑∫
∑∑∫∑∫ ∑∑∫∫
= == =
= == ==
∂
σ∂+σ
∂∂
=σ∂∂
=
σ=σ==⋅
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
)()(i j V j
ijiij
j
i
i j Viji
j
i j Sjiji
i S jjiji
i Sii
S
dVx
wxw
dVwx
dSnwdSnwdSvwdSvw
q-Force System
q -Force System
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 64
int
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
)(21
)(
)(21
)(
)(
WdVdVxw
xw
dVxw
dVxw
dVxw
dVxw
dVxw
dVfx
wdVxw
dVfwdVx
wxw
W
i j Vijij
i j Vij
i
j
j
i
i j Vij
i
j
i j Vij
j
i
j i Vji
i
j
i j Vij
j
i
i j Vij
j
i
i V ji
j
iji
i j Vij
j
i
i Vii
i j V j
ijiij
j
iext
δ=σε=σ∂
∂+
∂∂
=
σ∂
∂+σ
∂∂
=
σ∂
∂+σ
∂∂
=σ∂∂
=
+∂
σ∂+σ
∂∂
=
+∂
σ∂+σ
∂∂
=δ
∑∑∫∑∑∫
∑∑∫∑∑∫
∑∑∫∑∑∫∑∑∫
∑∫ ∑∑∑∫
∑∫∑∑∫
= == =
= == =
= == == =
= == =
== =
∫ ∑∑∫∫= =
σε=⋅+⋅S i j V
ijijV
dVdVdS3
1
3
1fwvw
3.2.2 Framed Structures
)(∑ ∫∫∫∫∑∫ τγ+σε=τγ+σε=σ⋅εe VVVVij V
ijij dVdVdVdVdVee
Internal virtual work by normal stress – bending moment
∫∫∫∫ ∫
∫ ∫∫∫∫
=−−===
==−−=σε
eeee
e
e
eeee
llll
A
l
AVVV
dxEIMMdx
EIMEI
EIMdx
dxwdEI
dxwddx
dxwddAEy
dxwd
dAdxydx
wddx
wdEdVydx
wddx
wdEdVydx
wdEydx
wddV
0002
2
2
2
02
22
2
2
0
22
2
2
22
2
2
2
2
2
2
2
2
)()()(
)()(
Internal virtual work by normal stress – Axial Force
∫∫ ∫∫∫ ===σεee
eee
ll
AVV
dxEA
FFdxAFdA
EAFdV
AF
EAFdV
00
)(
Internal virtual work by shear stress
QyIb
V)(
=τ and QyGIb
V)(
=γ where ∫=a
y
ydAQ
∫∫ ∫∫∫ ===τγee
eee
ls
l
AVV
VdxVGAf
dAdxybI
QGVVQdV
yIbVQ
yGIbVdV
0022
2
)()()(
Total displacement
∑ ∫ ∫∫ ++=e
l ll
s
e ee
dxEA
FFdxGA
VVfdxEIMMxw
0 000 )()(
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 65
3.2.3 Effect of Shear Deformation
For simple beam with a uniform load case
EIqllwM 384
5)2
(4
=
Shear Effect
▬ Shear force of load case q
▬ Shear force of load case q
▬ Deflection by shear force
GAqlf
VVlGA
fVdxV
GAf
VdxVGAf
w ssl
sl
sS
e
822122 22/
00
==== ∫∫
GAEI
lf
EIqlGAqlf
ww ss
M
s
40384
384/58/
24
2
==
for a rectangle section of bh× with steel
2
2
2
3
5.21240
6.238456
lh
bhlbh
ww
M
s =×
××=
For small h/l, the effect of shear deformation can be neglected.
1/2
ql/2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 66
3.3 Truss Problems
3.3.1 Principle of Virtual Work
From General principle
∑∑ ∫ ∫∫
∫
=++=
α+=α=⋅
ii
ii
e
l ll
s
iiiS
lEA
FFdx
EAFFdx
GAVVfdx
EIMM
vuwdS
e ee
0 00
22
)(
coscoswq
From equilibrium equation
0 , 0)(
1
)(
1=+−=+− ∑∑
==
iim
j
ij
iim
j
ij YVXH for njni ,,1=
where m(i), ijH and i
jV 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.
0])( )[(1
)(
1
)(
1=+−++−∑ ∑∑
= ==
njn
i
iiim
j
ij
iiim
j
ij vYVuXH
∑∑
∑∑ ∑∑
∑ ∑∑
==
== ==
= ==
+=−θ+−θ
+=θ+θ
=+θ−++θ−
njn
i
ii
ii
nmb
iiii
eiiii
ei
njn
i
iiiinjn
i
im
j
ij
ij
iim
j
ij
ij
i
njn
i
iiim
j
ij
ij
iiim
j
ij
ij
vYuXvvFuuF
vYuXFvFu
vYFuXF
11
1212
11
)(
1
)(
1
1
)(
1
)(
1
) ())(sin )(cos(
) ()sin cos(
0))sin( )cos((
∑∑
∑∑∑∑
==
====
+=
==∆=−θ+−θ
n
i
iiiinmb
i i
iie
i
nmb
i i
iie
i
i
iinmb
i
ei
nmb
ii
ei
nmb
iiiiiii
ei
vYuXEA
lFFEA
lFFEA
lFFlFvvuuF
11
1111
1212
) ()(
))(sin )((cos
iY iX
iF1−
ijF−
iimF )(−
iY iX
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 67
or, by applying Betti-Maxwell reciprocal theorem
∑∑==
=+nmb
i i
iie
injn
i
iiii
EAlFFvYuX
11 )() (
The displacement of a joint k in a truss is obtained by applying a unit load at a joint k in an arbitrary direction.
∑=
===+nmb
i i
iie
ikkkkkk
EAlFFvYuX
1 )(coscos αα uuX
Since α represnts the angle between the applied unit load and the displacement vector, αcosku are the displacement of the joint k in the direction of the applied unit load.
For vertical displacement For Horizontal displacement
ivv θ− sin)( 12
iuu θ− cos)( 12
iθ iθ
)( 12 uu −
)( 12 vv −
ku αcosku
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 68
3.3.2 Example
Real System Virtual System
Table for calculation of the deflection of a truss
Member EAl F F EA
lFF
1 1 30 0.75 22.5
2 2 -30 2 -0.75 2 45 2
3 1 30 0.75 22.5
4 1 40 0.50 20
5 2 -10 2 0.25 2 -5 2
6 1 -30 -0.75 22.5
7 1 20 0 0
8 1 40 0.5 20
9 2 -10 2 -0.25 2 5 2
10 1 -30 -0.25 7.5
11 1 30 0.25 7.5
12 1 30 0.25 7.5
13 2 -30 2 -0.25 2 15 2
∑ 130+60 2
EAl
EAL 215)260130( =+=δ
1
2
3
4
5
6
7
8
9
10
11
12
13
20 20 20
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 69
3.3.3 Conservation of Energy
Equilibrium and Conservation of Energy
▬ Equilibrium Equation
0 , 0)(
1
)(
1=+−=+− ∑∑
==
iim
j
ij
iim
j
ij YVXH for njni ,,1=
▬ External work
∑ ∑∑∑∑= ====
+=⋅∆=+=njn
ii
im
j
iji
im
j
ij
njn
iii
njn
iiiiiext vVuHvYuXW
1
)(
1
)(
111) (
21
21)(
21 P
int
nmb
i i
ie
ie
inmb
ii
ei
nmb
iiii
eiiii
ei
njn
i
im
j
ij
iji
im
j
ij
iji
njn
ii
im
j
ij
iji
im
j
ij
ij
njn
ii
im
j
iji
im
j
ijext
WEA
lFFlF
vvFuuF
FvFu
vFuFvVuHW
==∆=
−θ+−θ=
θ+θ=
θ+θ=+=
∑∑
∑
∑ ∑∑
∑ ∑∑∑ ∑∑
==
=
= ==
= === ==
11
1
1212
1
)(
1
)(
1
1
)(
1
)(
11
)(
1
)(
1
21
21
))(sin )(cos(21
)sin cos(21
)sin cos(21) (
21
intext WW =
Conservation of Energy in each load case
∑∑==
⋅∆=njn
iii
nmb
i i
ii
EAlF
11
2
21
21 P , ∑∑
==
⋅∆=njn
iii
nmb
i i
ii
EAlF
11
2
21
21 P
Two load cases are applied simultaneously
∑∑∑∑====
⋅+⋅=→+⋅+=+ njn
iiiii
nmb
i i
iinjn
iiiii
nmb
i i
iii
EAlFF
EAlFF
1111
2
)ΔΔ(21)()ΔΔ(
21)(
21 PPPP
External work for sequential loading (P first)
∑∑∑∑
∑∑∑∑
====
====
⋅∆=⋅∆→⋅∆=⋅∆+⋅∆
∆+⋅∆+⋅∆=+⋅∆+∆
njn
iii
njn
iii
njn
iii
njn
iiiii
njn
iii
njn
iii
njn
iii
njn
iiiii
1111
1111
)(21
21
21)()(
21
PPPPP
PPPPP
∑∑∑===
⋅∆=⋅∆=njn
iii
njn
iii
nmb
i i
ii
EAlFF
111PP
∑∑∑===
δ=δ=njn
iii
njn
iii
nmb
i i
ii lFlFEA
lFF111
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 70
3.4 Frame Problems
∑ ∫ ∫∫ ++=∆e
l ll
s
e ee
dxEA
FFdxGA
VVfdxEIMM
0 00
)(
where ∆ is the displacement in the direction of applied unit concentrate load in the virtual system.
Moment Shear Axial
+=δ ∫∫
2/
00
2 ll
M dxdxEI
EIPllPlllPll
EIM 16
3)
446443(2
=××+××=δ
l
l
HA HB=P/4
RA=P/2 RB=P/2
P
-
Pl/4
+
P/2
-
- +
P/4
- -
- P/2
P/4
×
Pl/4 l/4
×
Pl/4 l/4
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 71
+=δ ∫∫
2/
00
2 ll
sS dxdx
GAf
EAPlfPlPl
GAf
dxGA
VVf ss
V
sS 8
3)21
2241
4(
2=××+××==δ ∫
+= ∫∫
2/
00
2 ll
sA dxdx
EAδ
EAPlPlPl
EAdx
EAAA
VA 16
9)41
4221
2(2
=××+××==δ ∫
))(75.0)(56.11(16
)961(16
223
22
3
lh
lh
EIPl
EAlEI
GAlEIf
EIPl s
ASM
++=
++=δ+δ+δ=δ
In most cases, the deformation caused by the shear force and the axial force negligibly small compared to that caused by the bending moment. If this is the case, the displacement of a frame can be approximated by considering only the bending moment.
∑∫=∆e
le
dxEIMM
0
×
P/4 1/4
×
P/2 1/2
×
P/4 1/4
×
P/2 1/2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 72
This page is intentionally left blank.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 73
Chapter 4
Analysis of Statically Indeterminate Beams
1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 74
4.1 Propped Cantilever Beam
Equilibrium equation
02
0
=−+
=+−−
lRlqlM
qlRR
BA
BA
4.1.1. The first idea
=
+
Compatibility condition
00 =δ+δ xBR
– The end displacements of the cantilever beam for two loads cases are calculated by
the principle of virtual work.
EI
qllqllEI
dxEI
dxEIMM ll
8))(
2(
41 1 42
000 =−−===δ ∫∫
EIllll
EIdx
EIdx
EIMM ll
x 3))((
31 1 3
00
=−−===δ ∫∫
q
RA RB
MA
δ0
xBR δ
-ql2/2
1
-l
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 75
Compatibility condition and the final solution
(up)
Moment Diagram
Deflected shape
4.1.2. The second idea
=
+
Compatibility condition
-ql2/2
3ql2/8
-ql2/8 -
3l/8
+ =
θ0
MA
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 76
– Rotional Angle at the fixed end
EIqlqll
EIdx
EIdx
EIMM ll
241
831 1 32
000 =×===θ ∫∫
EIll
EIdx
EIdx
EIMM ll
x 311
31 1
00
=××===θ ∫∫
Compatibility condition and the final solution
23
810
324qlM
EIlM
EIql
AA −=→=+
Other reactions by a free body diagram
=
ql2/8 1
1
ql/2 ql/2 ql/8
ql2/8
ql/8
+
3ql/8 5ql/8
ql2/8
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 77
4.2 Cantilever Beam with Spring Support
Robin BC (The third type BC)
Primary structure
Compatibity Condition
wbeam(l)=δspring →
As , and As
Deflected Shape for
δ0
Ra
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 78
4.3 Support Settlement
Primary structure
Compatibility condition
Deflected Shape
∆
RB
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 79
4.4 Temperture Change
Primary structure
Curvature due to temperature change
For simple beam, →
Comaptibility condition
→ →
T1
T2
θ0
MA
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 80
4.5 Shear Effect
Primary Structures – Shear force diagram
Compatibility condition
0)()( 000 =δ+δ+δ+δ=δ+δ Sx
MxB
SMxB RR
GAfqlqll
GAfdx
GAfdx
GAVVf
llS
2)1)((
2
2
000 ====δ ∫∫
GAfll
GAfdx
GAfdx
GAVVf
llSx ====δ ∫∫ )1)(1(
00
2
2
2
2
3
24
)/(56.11)/(1.21
83
31
41
83
3
28lhlhql
GAlfEI
GAlfEI
ql
GAfl
EIql
GAfql
EIql
RB ++
−=+
+−=
+
+−=
– For retangular section
22
2
3
2 )(52.0)(125
6.212
)1(2
1256
lh
lh
hblv
E
bhE
GAlfEI
=××
=
+
=
For 201
=lh
830014.1
039.010053.01
83
)/(56.11)/(1.21
83
2
2 qlqllhlhqlRB −=
++
−=++
−= (0.14 % error)
1
ql 1
+ +
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 81
For 101
=lh
830053.1
0156.01021.01
83
)/(56.11)/(1.21
83
2
2 qlqllhlhqlRB −=
++
−=++
−= (0.53 % error)
For 51
=lh
8302.1
0624.01084.01
83
)/(56.11)/(1.21
83
2
2 qlqllhlhqlRB −=
++
−=++
−= (2.0 % error)
You may neglect the effect of the shear deformation in most cases !!
4.6 2-Span Continuous Beam
Primary structure
Compatibility
)( 00RxB
RLxB
L MM θ+θ−=θ+θ → 0)(00 =θ+θ+θ+θ Rx
LxB
RL M
EI EI q
ql
Rxθ L
xθ
1
4
2ql 8
2ql
q
ql
R0θ L
0θ
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 82
Deflected shape
1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 83
4.7 Fixed-Fixed End Beam
4.7.1. Primary Structure type I
EIqlqlll
EIdx
EIMMl
8)
2()(
41 42
0
0110 =−×−×==δ ∫
EIqlqll
EIdx
EIMMl
6)
2(1
31 32
0
0220 −=−××==δ ∫
EIllll
EIdx
EIMMl
3)()(
31 3
0
1111 =−×−×==δ ∫
EIlll
EIdx
EIMMl
21)(
21 2
0
212112 −=×−==δ=δ ∫
EIll
EIdx
EIMMl
=××==δ ∫ 111
0
2222
q
RA RB
MA MB
-ql2/2
M0
1
-l
M1
1 1
M2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 84
Compatibility condition (Flexibility equation)
=+−−
=−+→
=δ+δ+δ=δ+δ+δ
026
0238
00
21
23
2
2
1
34
22212120
21211110
XEIlX
EIl
EIql
XEIlX
EIl
EIql
XXXX
21qlX −= ,
12
2
2qlX −=
4.7.2. Primary Structure type II
EIqlqll
EIdx
EIMMl
2481
31 42
0
0110 =××==δ ∫
EIqlqll
EIdx
EIMMl
2481
31 32
0
0220 =××==δ ∫
EIll
EIdx
EIMMl
3)1()1(
31
0
1111 =−×−×==δ ∫
EIll
EIdx
EIMMl
611
61δδ
0
212112 =××=== ∫
EIll
EIdx
EIMMl
311
31
0
2222 =××==δ ∫
M0
M2
1
1
M1
1
1 8
2ql
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 85
Compatibility condition (Flexibility equation)
Reactions and Moment Diagrams
Deflected Shape
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 86
4.8 3-Span Continuous Beam
4.8.1. Uniform load case
Primary structure
)1(248
13
18
13
1
2
1
1
32
2
2
1
2
0
0110 I
IEI
qlqllEI
qllEI
dxEIMMl
+=××+××==δ ∫
)1(248
13
18
13
1
2
1
1
32
1
2
2
2
0
0220 I
IEI
qlqllEI
qllEI
dxEI
MMl
+=××+××==δ ∫
)1(3
113
1113
1
2
1
121
2
0
1111 I
IEIll
EIl
EIdx
EIMMl
+=××+××==δ ∫
220
212112 6
116
1δδEIll
EIdx
EIMMl
=××=== ∫
)1(3
113
1113
1
2
1
112
2
0
2222 I
IEIll
EIl
EIdx
EIMMl
+=××+××==δ ∫
EI1 EI2
q
EI1
M0
1
M1
1
M2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 87
Compatibility condition (Flexibility equation)
In case ,
4.8.2. Complicated Load Case
Primary structure
M0
EI EI
q
EI
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 88
EIqlqll
EIdx
EIMMl
2481
31 32
0
0110 =××==δ ∫
EIql
EIql
EIqlqll
EIqll
EIdx
EIMMl
485
162441)
211(
61
81
31 333222
0
0220 =+=××++××==δ ∫
Compatibility condition (Flexibility equation)
=++
=++→
=δ+δ+δ=δ+δ+δ
032
6485
063
224
00
21
3
21
3
22212120
21211110
XEIlX
EIl
EIql
XEIlX
EIl
EIql
XXXX
2
1 401 qlX −= , 2
2 406 qlX −=
Compatibility Condition (Flexibility Equation) in General
∑=
∆=δ+δn
jijiji X
10
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 89
Chapter 5
Analysis of Statically Indeterminate Trusses
1 1
1
21
− 21
−
21
−
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 90
5.1 Various Types of Trusses
Determinate Truss
Externally Indeterminate Truss
Internally Indeterminate Truss
Mixed Indeterminate Truss
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 91
5.2 A Simple Truss
5.2.1 Method - I
+
P
X
P
P
=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 92
Primary structure
At L1
075.06.0
25.118.0
1
31
3
==+
==
FFF
FF
P -1.0
1.25
0
-0.75 0
1 0.75 0.75
F1 F3
0.75
1
①,②,④:0.5A ③,⑤:A
P ②
①
③
④
⑤
L
0.75L
0.75L L1 L2
U1 U2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 93
At L2
75.0 , 25.1
18.006.0
45
5
54
−===
=+
FFF
FF
At U1
1 , 75.0
08.006.0
21
52
51
−=−==+=+
FFFFFF
1
-1.0
1.25
1.25
-0.75
1
-0.75
F4 F5
1
F1
F2
F5
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 94
Axial force table for primary structure
Mem 0F xF A L LEA
FF x00 =δ L
EAFx
x
2
=δ
1 0 -0.75 0.5 0.75L 0 LEA5.0
75.0 3
2 -P -1.0 0.5 L EAPL5.0
LEA5.0
1
3 1.25 P 1.25 1.0 1.25L EAPL325.1
EAL325.1
4 -0.75 P -0.75 0.5 0.75L EAPL
5.075.0 3
LEA5.0
75.0 3
5 0 1.25 1.0 1.25L 0 EAL325.1
∑ EAPL79.4
EAL59.7
Compatibility Condition
00 =δ+δ xX PX 63.0−=→ Final Solution
Mem 0F xF xXF xXFF +0
1 0 -0.75 0.47P 0.47P
2 -P -1.0 0.63P -0.37P
3 1.25 P 1.25 -0.79P 0.46P
4 -0.75 P -0.75 0.47P -0.28P
5 0 1.25 -0.79P -0.79P
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 95
5.2.2 Method - II
P
+
P
X
X
=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 96
Compatibility condition
AEXLX x −=δ+δ0
Primary structure
-1.0P P
1.25P -0.75P
P
0
0.75P 0.75P
-0.8
1.0 -0.6
0.8
-0.6
0.8
1
1
X
= +
F0 Fx
AEXL
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 97
Axial force table for primary structure
Mem 0F xF A L LEA
FF x00 =δ L
EAFF xx
x =δ
1 0 -0.6 0.5 0.75L 0 EAL54.0
2 -P -0.8 0.5 L EA
PL6.1 EAL28.1
3 1.25 P 1.0 1.0 1.25 L EA
PL56.1 EAPL25.1
`4 -0.75 P -0.6 0.5 0.75 L EAPL68.0
EAPL54.0
5 - - 1.0 - - -
∑ EAPL84.3 EA
L61.3
AEXLX x −=δ+δ0 →
AEXLX
AEL
AEPL 25.161.384.3 −
=+
PPX 79.086.484.3
−==
PXH 63.08.02 −==
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 98
5.3 A Truss with 1 Roller Support
Primary Structures
1
2 3
4
5
6
7
8
9
10
3P
P 2P 3P
-P
2P P 2P
P 3P - 2 P - 2 P - 22 P
1
1
1
21
− 2
1−
21
−
21
−
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 99
Axial force table for primary structure
Mem F 0 xF
EAL L
EAFF x0
0 =δ LEA
FF xxx =δ
1 P 0 1 0 0
2 P2− 0 2 0 0
3 P 21
− 1 2P
− 21
4 P2 21
− 1 22P
− 21
5 P2− 1 2 P2− 2
6 - - 2 - -
7 P− 21
− 1 2P
21
8 P3 21
− 1 23P
− 21
9 P2 0 1 0 0
10 P22− 0 2 0 0
-5.54P 3.41
Compatibility condition
LEAXXx 20 −=δ+δ
PXXXP 15.141.141.354.5 =→−=+−
X
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 100
Temperature change and fabrication error
)22(0 fx LTLEAXX ∆+∆α+−=δ+δ
In case of no external loads
)2(21.0)2(82.4 ff LTL
EAXLTXEAL
∆+∆−=→∆+∆−= αα
5.4 Truss with Two Hinge Supports
Primary structure and compatibility condition
0
2
22212120
121211110
=δ+δ+δ
−=δ+δ+δ
XX
LEAX
XX
3P 2P P
X1
1
2 3
4
5
6
7
8
9
X1
X2
X1
X1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 101
– F0
– F1
– F2
1
1
1 1 1
2P P 3P
-P
2P 2P P
P 3P - 2 P - 2 P - 22 P
21
−
1 1
1
21
− 21
−
21
−
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 102
Axial force table for the primary structure
Mem 0F 1F 2F EAL L
EAFF 10 L
EAFF 20 L
EAFF 11 L
EAFF 12 L
EAFF 22
1 P2 0 1 1 0 P2 0 0 1
2 P22− 0 0 2 0 0 0 0 0
3 P3 22
− 0 1 P2
23− 0 2
1 0 0
4 P2 22
− 1 1 P2
22− P2 2
1 22
− 1
5 P2− 1 0 2 P2− 0 2 0 0
6 P− 22
− 0 1 P22 0 2
1 0 0
7 P 22
− 0 1 P22
− 0 21 0 0
8 P 0 1 1 0 P 0 0 1
9 P2− 0 0 2 0 0 0 0 0
∑ -5.54P 5P 3.41 -0.71 3
0371.0541.171.041.354.5
21
121
=+−−=−+−
XXPXXXP
® PXX
PXX5371.0
54.571.082.4
21
21
−=+−=−
® PX
PX44.1
94.0
2
1
−==
Structural Analysis Lab. http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 103
Chapter 6
Analysis of Statically Indeterminate Frames
EI,l
w
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 104
6.1 Γ-shaped Frame-I
Equilibrium equation
02
0
=−−
=++
lRPlM
PRR
CA
CA
6.1.1 Primary Structure type I
+
δ0
P 1
Rcδx
P
RA
MA
RC l
0.75L
l
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 105
Compatibility condition
0δ δ0 =+ xCR
End Displacements
333
2
0000
4829)
485
2(1)}
2)(2
2(
61
2))(
2({1
1 1
PlEI
PlPlEI
PlllllPllEI
dxEI
dxEI
dxEIMM
lll
=+=−−−×+−−=
+==δ ∫∫∫
333
000
34)
31(1)})((
3))(({1
1 1
lEI
llEI
llllllEI
dxEI
dxEI
dxEIMM lll
x
=+=−−+−−=
+==δ ∫∫∫
Compatibility condition and the final solution
03
448
29 33 =+ lREI
PlEI C
PPRC 45.06429
−=−= , PPRA 55.06435
−=−= , PlM A 643
−=
P
-
-
l
1
l
-
Pl/2
P
Pl/2
- 1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 106
Moment Diagram
+
=
Deflection Shape
Pl/2
P
Pl/2
-
-
P
0.45P
+
0.45Pl
0.45P
0.45Pl
+
0.225Pl
+
-
- 0.05Pl
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 107
6.1.2. Primary Structure type II
Compatibility Condition
00 =θ+θ xBM Rotation Angle
EIPlPll
EIdx
EIdx
EIMM ll
161
423
61 1 2
000 =××××===δ ∫∫
EIlll
EIdx
EIdx
EIMM ll
x 34)
3(1) (1
0
22
0
=+=+==δ ∫∫
P
0θ
P
P/2
P/2
Pl/4
+
1
+
+ 1/l
1/l
MB
xBM θ
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 108
Compatibility condition and the final solution
034
16
2
=+ BMEIl
EIPl →
643PlM B −=
6.2 Γ-shaped Frame-II
6.2.1 Primary Structure type I
w
RA
MA
RB
EI,l
HB
EI,l
-wl2/2
M0 -
l
M1
1
+
+
M2
1
l
+
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 109
EIwllwll
EIdx
EIdx
EIMM ll
6))(
2(
31 1 42
00
0110 −=−===δ ∫∫
EIwllwll
EIdx
EIdx
EIMM ll
8))(
2(
41 1 42
00
0220 −=−===δ ∫∫
EIlllllll
EIdx
EIdx
EIMM ll
34)})((
3))(({1) (1 3
0
22
0
1111 =+=+==δ ∫∫
EIllll
EIdx
EIdx
EIMM ll
2))((
21 1 3
00
212112 ====δ=δ ∫∫
EIllll
EIdx
EIdx
EIMM ll
3))((
31 1 3
0
2
0
2222 ====δ ∫∫
Compatibility condition (Flexibility Equation)
=++−
=++−→
=δ+δ+δ=δ+δ+δ
0328
023
46
00
2
3
1
34
2
3
1
34
22212120
21211110
XEIlX
EIl
EIwl
XEIlX
EIl
EIwl
XXXX
281wlX −= ,
73
2wlX =
Reactions
w
wl/28
4wl/7
3wl2/28
wl/28
3wl/7
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 110
Moment Diagram
Deflected Shape
6.2.2. Primary Structure type II
-3wl2/28
11wl2/196
-wl2/28
M0
wl2/8
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 111
EIwldx
EIdx
EIMM ll
24 1 3
00
012010 ===δ=δ ∫∫
EIll
EIdx
EIdx
EIMM ll
3)1)(1(
31 1
0
2
0
1111 ====δ ∫∫
EIldx
EIdx
EIMM ll
6 1
00
212112 ===δ=δ ∫∫
EIldx
EIdx
EIMM ll
32) (1
0
22
0
2222 =+==δ ∫∫
Compatibility condition (Flexibility Equation)
=++
=++→
=δ+δ+δ=δ+δ+δ
032
624
06324
00
21
3
21
3
22212120
21211110
XEIlX
EIl
EIwl
XEIlX
EIl
EIwl
XXXX
28
2
1wlX −= ,
283 2
2wlX −=
M1
1
+
M2
1
+
+
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 112
6.3 Portal Frame subject to Horizontal Load
6.3.1. Primary Structure type I
1
w
M0
-wl2/2
-
EI,l
w
M2
l
+
+
M3
1
1
+
+
+
M1
-l
- -
-
1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 113
EIwldx
EIdx
EIMM ll
24 1 4
00
0110 ===δ ∫∫
EIwldx
EIdx
EIMM ll
6 1 4
00
0220 −===δ ∫∫
EIwldx
EIdx
EIMM ll
6 1 3
00
0330 −===δ ∫∫
EIldx
EIdx
EIMM ll
35) (21 3
0
22
0
1111 =+×==δ ∫∫
EIldx
EIdx
EIMM ll 3
00
212112 21
−=×==δ=δ ∫∫
EIldx
EIdx
EIMM ll 2
0
2
0
313113
2) (21−=+×==δ=δ ∫∫
EIldx
EIdx
EIMM ll
34) (1 3
0
22
0
2222 =+==δ ∫∫
EIldx
EIdx
EIMM ll
23 ) (1 2
0
2
0
323223 =+==δ=δ ∫∫
EIldx
EIdx
EIMM ll 3 3 1
0
2
0
3333 =×==δ ∫∫
Compatibility condition (Flexibility Equation)
→
=δ+δ+δ+δ=δ+δ+δ+δ
=δ+δ+δ+δ
00
0
33323213130
32322212120
31321211110
XXXXXX
XXX
=++−−
=++−−
=−−+
03232
6
023
34
6
0235
24
32
2
1
23
3
2
2
3
1
34
3
2
2
3
1
34
XEI
lXEIlX
EIl
EIwl
XEIlX
EIlX
EIl
EIwl
XEIlX
EIlX
EIl
EIwl
Matrix form
−
−−=
−
−
−−
EIwlEI
wlEI
wl
XXX
EIl
EIl
EIl
EIl
EIl
EIl
EIl
EIl
EIl
6
6
24
3232
23
34
235
3
4
4
3
2
1
22
233
233
→
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Dept. of Civil and Environmental Eng., SNU 114
Reactions
Moment Diagram
Deflected Shape
w
5wl/24 19wl/24
wl /7 wl /7
59wl2/252 31wl2/252
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 115
6.3.2. Primary Structure type II
EIwldx
EIdx
EIMM ll
6 1 3
00
0110 −===δ ∫∫
EIwldx
EIdx
EIMM ll
8 1 3
00
0220 ===δ ∫∫
EIwldx
EIdx
EIMM ll
8 1 3
00
0330 −===δ ∫∫
EIldx
EIdx
EIMM ll
34) (1
0
22
0
1111 =+==δ ∫∫
1
1 +
+ M1
+
+
-
1
M2
+ +
1
1
M3
M0 w
-wl2/2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 116
2100
2112 3
) (1δ=−=+==δ ∫∫ EI
ldxEI
dxEIMM ll
3100
3113 2
1δ====δ ∫∫ EI
ldxEI
dxEIMM ll
EIldx
EIdx
EIMM ll
=×==δ ∫∫0
2
0
2222 31
3200
3223 6
) (1δ=−=+==δ ∫∫ EI
ldxEI
dxEI
MM ll
EIldx
EIdx
EIMM ll
32 2 1
0
2
0
3333 =×==δ ∫∫
Compatibility condition (Flexibility Equation)
00
0
33323213130
32322212120
31321211110
=δ+δ+δ+δ=δ+δ+δ+δ
=δ+δ+δ+δ
XXXXXX
XXX
Matrix Form
−
−
−=
−
−−
−
EIwlEI
wlEI
wl
XXX
EIl
EIl
EIl
EIl
EIl
EIl
EIl
EIl
EIl
8
8
6
32
62
63
2334
3
3
3
3
2
1
−=
−
−
−
−
−=
−
−
−
−−
−
−=
−
25231
50443
50429
8
8
6
2144
212
2116
212
2123
215
2116
215
2123
8
8
6
32
62
63
2334
2
3
3
3
3
3
31
3
2
1
wl
EIwlEI
wlEI
wl
lEI
EIwlEI
wlEI
wl
EIl
EIl
EIl
EIl
EIl
EIl
EIl
EIl
EIl
XXX
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 117
6.4 Portal Frame subject to Vertical Load
1
1 +
+ M1
+
+
-
1
M2
+ +
1
1
M3
EI,l
a P
M0
4Pab
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 118
Matrix Form
Sidesway :
Deflected Shape
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 119
6.5 Order of Indeterminacy
# of unknowns
# of member × # of internal force per member +
# of reactions - # of known quantities
# of equations
# of member × # of E.E. per member +
# of joints × # of E.E. per joint - # of used equations
# of Indeterminacy = # of unknowns - # of equations
Order of Indeterminacy of the frame shown above
6×10 +3×3 – (3×10 +3×9) = 69 – 57 = 12
6.5.1. Order of Indeterminacy – Beam
Number of Internal Forces in a Member : 4
Number of Equilibrium Equations in a Member : 2
Number of Equilibrium Equations at a Joint : 2
Simple Beam
- # of unknowns : 4×1 + 1×2 = 6 or 4×1 + 2×2 – 2= 6
- # of equations : 2×1 + 2×2 = 6
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 120
- Order of Indeterminacy : 6 – 6 = 0
Overhanged Beam
- # of unknowns : 4×2 + 1×2 – 2 = 8
- # of equations : 2×2 + 2×2 = 8
- Order of Indeterminacy : 8 – 8 = 0
Gerber Beam
- # of unknowns : 4×3 + 1×3 – 2 = 13
- # of equations : 2×3 + 2×4 – 1= 13
- Order of Indeterminacy : 13 – 13 = 0
Continuous Beam
- # of unknowns : 4×3 + 1×4 = 16
- # of equations : 2×3 + 2×4 = 14
- Order of Indeterminacy : 16 – 14 = 2
6.5.2. Order of Indeterminacy - Truss
Number of Internal Forces in a Member : 1
Number of Equilibrium Equations in a Member : 0
Number of Equilibrium Equations at a Joint : 2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 121
Determinate Truss
- # of unknowns : 1×21 + 3 = 24
- # of equations : 2×12 = 24
- Order of Indeterminacy : 24 – 24 = 0
Internally Indeterminate Truss
- # of unknowns : 1×25 + 3 = 28
- # of equations : 2×12 = 24
- Order of Indeterminacy : 28 – 24 = 4
Internally Indeterminate Truss
- # of unknowns : 1×25 + 5 = 30
- # of equations : 2×12 = 24
- Order of Indeterminacy : 30 – 24 = 6
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 122
6.5.3. Order of Indeterminacy - Frame
Number of Internal Forces in a Member : 6
Number of Equilibrium Equations in a Member : 3
Number of Equilibrium Equations at a Joint : 3
Internally Indeterminate Frame
- # of unknowns : 6×3 + 3 = 21
- # of equations : 3×3 + 3×4 = 21
- Order of Indeterminacy : 21 – 21 = 0
Frame with Hinges
- # of unknowns : 6×4 + 4 – 2 = 26
- # of equations : 3×4 + 3×5 – 1 = 26
- Order of Indeterminacy : 26 – 26 = 0
Portal Frame with Fixed Supports
- # of unknowns : 6×3 + 6 = 24
- # of equations : 3×3 + 3×4 = 21
- Order of Indeterminacy : 24 – 21 = 3
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 123
6.5.4. Selecting a Primary Structure for a Complicated Frame
Order of Indeterminacy of the frame
- # of unknowns : 6×10 + 3×3 = 69
- # of equations : 3×10 + 3×9 = 57
- Order of Indeterminacy : 69 – 57 = 12
By Replacing a Rigid Joint with a Hinge, we can reduce
- # of unknowns by the number of members at the joint
- # of equations by one
In the Primary Structure
- Reduction in unknowns : 1 + 1 + 3 + 4 + 3 + 2 + 3 = 19
- Reduction in equations : 5
- Reduction in Indeterminacy : 17 – 5 = 12
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 124
6.6 General Frame
Primary Structure
M1, M2, M3
EI
L
L/2
2EI
q qL
EI
L L
EI
ql2/8 4
2ql M0
M1 M2
M3
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 125
EIqlql
EIldx
EIMMl
2481
3
32
0
0110 =××==δ ∫
EIqlqlldx
EIMMl
164)1()
211(
6
32
0
0220 −=×−×+==δ ∫
00
0330 ==δ ∫
l
dxEIMM
EIldx
EIMMl
30
1111 ==δ ∫ , 01312 =δ=δ
EIldx
EIMMl
30
2222 ==δ ∫ , 02321 =δ=δ
EIldx
EIMMl
620
3333 ==δ ∫ , 03231 =δ=δ
ql2/8 4
2ql
+ 1
- 1
+ 1
M0 M1
M2 M3
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 126
Compatibility Condition
321 δ=δ=δ
1
3
313212111101 324M
EIl
EIqlMMM +=δ+δ+δ+δ=δ
2
3
323222121202 316M
EIl
EIqlMMM +−=δ+δ+δ+δ=δ
3333232131303 6M
EIlMMM =δ+δ+δ+δ=δ
One Additional Equilibrium Equation
0321 =++ MMM → 213 MMM −−=
Final Compatibility Condition
062246324 21
3
31
3
=++=−+ MEIlM
EIl
EIqlM
EIlM
EIl
EIql
026166316 21
3
32
3
=++−=−+− MEIlM
EIl
EIqlM
EIlM
EIl
EIql
21 64
9 qlEI
M −= , 22 64
11 qlEI
M = , 23 64
2 qlEI
M −=
In case n members are connected to a joint, and a hinge is used to release moment at the
joint you, have n-1 compatibility equations and one equilibrium equation, which leads to
total of n-1 compatibility equations with n-1 unknowns.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 127
6.7 General Joint Compatibility
Compatibility Condition
nδ==δ=δ 21
j
k
njijj
n
jijii MM ∑∑
+==
δ+δ+δ=δ11
0 ni 1for =
One Additional Equilibrium Equation
021 =+++ nMMM
Joint i
M1
Mn
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 128
This page is intentionally left blank.
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Dept. of Civil and Environmental Eng., SNU 129
Chapter 7
Influence Lines for
Determinate Structures
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 130
7.1 Influence Function
Influence function
Convolution integral – Superposition
ξξξ dqxIxdR pp )(),()( =
∫=l
pp dqxIxR0
)(),()( ξξξ Dirac delta functions
0lim→ε
= = )( ξ−δ x
1221lim)0
210(lim)(
00
00
=εε
=+ε
+=ξ−δ→ε
ε+ξ
ε+ξ
ε−ξ
ε−ξ
→ε ∫∫∫∫ll
dxdxdxdxx
)()(2
)()(lim)(21lim
)0)(21)(0)((lim)()(
00
00
0
ξ=ξ′=ε
ε−ξ−ε+ξ=
ε=
+ε
+=ξ−δ
→ε
ε+ξ
ε−ξ→ε
ε+ξ
ε+ξ
ε−ξ
ε−ξ
→ε
∫
∫∫∫∫
fFFFdxxf
dxxfdxxfdxxfdxxxfll
ξ
∞
ξ
ε21
2ε
1 ξ
I(xp,ξ)
xp
ξ q(x)
R(xp)
q(ξ)dξ
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 131
Concentrated loads of intensity P at )( ξ−δ=ξ xP
Responses by several concentrated loads
∑
∑∫∫ ∑∫
=
==
ξ=
ξξ−ξδξ=ξξ−ξδξ=ξξξ=
n
iipi
n
i
l
iip
l n
iiip
l
pp
xIP
dPxIdPxIdqxIxR
1
1 00 10
),(
)(),()(),()(),()(
7.2 Influence Line for Simple Beams
7.2.1 Moment
2
0 l≤ξ≤
2
0)2
(12
ξ=→=ξ−×−−
ξ−xx MlMl
ll
P1 ξ1
I(xp,ξ)
ξn
Pn
RA=(l - ξ)/l
ξ
RB=ξ/l
P = 1
RA=(l - ξ)/l
ξ P = 1 Mx
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 132
ll≤ξ≤
2
2
02
ξξ −=→=−
− lMMll
lxx
Influence line
7.2.2. Shear Force
2
0 l≤ξ≤
lVV
ll
xxξ
−=→=++ξ−
− 01
RA=(l - ξ)/l
ξ P = 1
Vx
RA=(l - ξ)/l
ξ
RB=ξ/l
P = 1
+ L/4
RA=(l - ξ)/l
Mx
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 133
ll≤ξ≤
2
llVV
ll
xxξξ −
=→=+−
− 0
Influence line
7.2.3 Maximum Moment in a Simple Beam
010
≤ξ≤−L
20)()
10(
21)( max
PLxMLPxM pp =→+= ξ
104
1020 LLL
=−≤ξ≤
104at
207)(
2043)
10(
21
22)( max
LPLxMPLPLPPxM pp =ξ=→+ξ
=+ξ+ξ
=
105
104 LL
≤ξ≤
RA=(l - ξ)/l
Vx
+
1/2
+
P/2 ξ
P
L/10
2)2/( ξ
=LIM
22)2/( ξ
−=LLI M
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 134
104at
207)(
209
4))
10(
21
2(
22)( max
LPLxMPLPLLPPxM pp =ξ=→+ξ
−=+ξ−+ξ
=
109
105 LL
≤ξ≤
105at
4013)(
2014
43))
10(
21
2()
22(
2)( max
LPLxMPLPLLPLPxM pp =ξ=→+ξ
−=+ξ−+ξ
−=
7.3 Influence Line of a Gerber Beam
7.3.1. Shear Force at x= L/2
2
0 L≤ξ≤
LL≤ξ≤
2
LL 5.1≤ξ≤
Influence line
ξ′
Lξ ′2
Lξ ′
−21
Lξ ′
−21
L/2 L
RA=1
ξ P=1
Vx=0
RA=1
Vx=1
+
1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 135
7.3.2. Moment at the fixed end
L≤ξ≤0
LL 5.1≤ξ≤
Influence line
7.3.1 Maximum Moment in the Gerber Beam
∫∫++
==4/4/
)()(L
M
L
M dxxIqdxxqIMξ
ξ
ξ
ξ
– L430 ≤ξ≤
)4
2(84
)4
(21 LqLLLqM +ξ−=+ξ+ξ×−= → 22
max 2188.0327 qLqLM −=−=
ξ ξ−=xM
)2( ξ′−−= LM x
ξ′
Lξ′2
Lξ′
−21
-
L )23()2( ξ−−=ξ′−− LL −ξ
- L
L/4
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 136
– LL ≤ξ≤43
)358
13(2
)27
232
821(
2
)43)(2
25(
2))((
2222222 ξ−ξ+−−=ξ+ξ+ξ−−ξ−−=
−ξξ−+−ξ−ξ+−=
LLqLLLLq
LLLqLLqM
LLqM650)65(
2=ξ→=ξ−−=′
2
2222max
2292.014433
72150300117
2))
65(3
655
813(
2qL
qLqLqLLLLqM
−=
−=−+−
−=−+−−=
– LL 25.1≤ξ≤
22max 1875.0
163)4
23(
84)2
22(
2qLqLMLqLLLLqM −=−=→ξ′−−=ξ′−+ξ′−−=
– For 025.0 ≤ξ≤− L or LL 5.125.1 ≤ξ≤
The maximum moment should be smaller than any of the above cases. Therefore, the maximum moment is
2max 2292.0 qLM −= at L
65
=ξ
)2( ξ′−− L )22
( ξ′−−L
)225( ξ−− L −ξ
ξ−L LLL43)
4( −ξ=−+ξ
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 137
7.4 Indirect Load
Equivalent to
10 =+→=∑ bav RRF
0)22
()22
)(1(0 0
0
0
0
=−++−−→=∑ lRlllxll
lxM ba
lx
lllRb +
−=
20 ,
lx
lll
RR ba −+
=−=2
1 0
In case the unit load is applied directly on the simple beam,
10 =+→=∑ bav RRF
0)22
(0 0 =−+−→=∑ lRxllM ba
Ra Rb
l0
1-x/l0 x/l0
l Ra Rb
x
l0
l/2
P =1
l
Ra Rb
x (l-l0)/2
l/2
P =1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 138
lx
lllRb +
−=
20
,
lx
lllRR ba −
+=−=
21 0 (Statically equivalent to the indirect load)
0)2
)(2
()1(0 00
0
=ξ+−
−+
+ξ−−−→=∑ ξ
lllx
lll
lxMM
ξ−−
+−
−+
=ξ+−
−+
+ξ−−= )12(22
)2
()2
)(2
()1(0
00000
0 lx
lllll
lx
lllll
lx
lll
lxM
7.4.1 Influence line at the mid-span
422222)12(
22)
2( 000000
0
000 llllllll
llll
lx
lllll
lx
lll
M−
=−
−−+
=−−
+−
−+
=
7.4.2 Truss Case
Ra Rb
1-x/l0 x/l0
ξ
Ra Rb
1-x/l0 x/l0
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 139
Raax
61−=
Rbax
6=
x
7.5 Influence Line of Truss
7.5.1 Diagonal Member
ax ≤≤0
axF
axF
6201
61
22
−=→=−−+− axa 2≤≤
)651(202
61
22
axF
axa
axF −−=→=
−−−+−
axa 62 ≤≤
)6
1(206
122
axF
axF −=→=−+−
x a
xaa
ax −=
−−
21
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 140
Influence line for the diagonal member
axFax
62 0 −=≤≤
)651(2 2
axFaxa −−=≤≤
)6
1(2 62axFaxa −=≤≤
7.5.2. Bottom Member
Take moment about point A (clockewise +).
ax ≤≤0
axFFaxaa
ax
650)(1)
61( =→=−−×−×−
axa 6≤≤ )
61(0)
61(
axFFaa
ax
−=→=−×−
+
-
264
2
61
axRa 6
1−=
A
65
+
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 141
Chapter 8
Influence Lines for Indeterminate Beams
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 142
8.1 Influence Lines at Supports
8.1.1. Reaction Force
By the Flexibility Method
– Compatibility Condition
bb
bbbbbb d
dRddR ξ
ξ −=→=+× 0
– Betti-Maxwell’s Reciprocal Theorem
ξξξξ ==→−δ=ξ−δ ∫∫ bLb
L
x
L
xb ddddxdLxdxdx2
0
2
0
)()(
– Influence Line : bb
b
bb
bb d
ddd
R ξξ −=−=
Moment Diagram
EIL
EILLL
EILMM
EILdbb 48
)2(6223
2322
33
==×=×=
L L
ξ
Rb = ??
P = 1
ξ P = 1
dxξ
dbξ
P = 1
dxb
dbb
P = 1 P(2L)/4= L /2
1/2 1/2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 143
Calculation of Deflection
baxxEI
wxMwEI ++−=→−=−=′′12
121 3
– Boundary conditions
EILaaL
EILw
bw
40
410)(
00)0(22
=→=+−→=′
=→=
– Deflection of the Beam
)3(12
1 23 xLxEI
wdxb +−==
Influence Line
)](3)[(21
6/)3(
121 3
323
Lx
Lx
EILxLx
EIdd
Rbb
bb −=+−−=−= ξ
8.1.2. Moment
By the force method
1
ξ P = 1
θbξ
dxξ , θxξ
L
EI
L L
ξ P = 1
EI EI
M = 1 dxb
θbb
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 144
– Compatibility Condition
bb
bbbbbb MM
θθ
−=→=θ+θ× ξξ 0
– Betti-Maxwell’s Reciprocal Theorem
ξξξξ θ=θ=→θ−δ=ξ−δ ∫∫ bLb
L
x
L
xb ddxLxdxdx3
0
3
0
)()(
– Influence Line : bb
b
bb
bb
dM
θ−=
θθ
−= ξξ
Calculation of Deflection
i) Left span
baxLEIxw
LxMwEI LL ++−=→−=−=′′
6
3
– Boundary conditions
EILaaL
EILLw
bw
L
L
60
60)(
00)0(2
=→=+−→=
=→=
– Deflection of the left span
)(6
1 23 xLxLEI
wd Lxb +−==
EILL
bb 3−=θ (counterclockwise)
ii) Analysis of Center and Right Span
M = 1 dxb
θbb
M = 1 dxb
M = 1
x = 1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 145
EIL
cc 32
=θ , EIL
cb 6=θ
– Compatibility condition: 410 −=
θθ
−=→=θ+θcc
cbcccccb MM
– Moment Diagram
iii) Deflection of Center span
baxxxLEI
wxL
MwEI cc ++−=→+−−=−=′′ )224
5(1)145(
23
EILaaLLL
EILw
bw
c
c
2470)
2245(10)(
00)0(22
=→=+−→=
=→=
)7125(24
1 223 xLLxxLEI
wcxb +−==δ
EILwc
Rbb 24
7)0( =′=θ (Clockwise)
EIL
EIL
EILR
bbLbbbb 8
5247
3=+=θ+θ=θ
iv) Deflection of Right Span
baxxL
xEI
wLxMwEI RR +++−=→−−=−=′′ )
824(1)
41
4(
23
EILaaLLL
EILw
bw
R
R
120)
824(10)(
00)0(22
−=→=++−→=
=→=
)23(24
1 223 xLLxxLEI
wRxb +−−==δ
Final Influence Line
i) left span :
)(15
485/)(
61 23
223 xLx
LEILxLx
LEIwM
bb
Lb −=+−−=
θ−=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 146
LLxLxL
Mb 577.0310)3(
154 22
2 ==→=−=′
LLLMb 103.0)1577.0)(1577.0(577.0154)577.0( −=−−=
ii) Center Span :
))(75(15
)7125(15
185/)7125(
241
2223
2
223
LxLxLxxLLxx
L
EILxLLxx
LEIwM
bb
Cb
−−−=+−−=
+−−=θ
−=
LLxLxLxL
Mb 384.015
39120)72415(15
1 222 =
−=→=+−−=′
LLLMb 080.0)1384.0)(7384.05(15384.0)384.0( −=−−×−=
iii) Right Span:
))(2(15
1)23(15
185/)23(
241
2223
2
223
LxLxxL
xLLxxL
EILxLLxx
LEIwM
bb
Rb
−−=+−=
+−=θ
−=
LxLxLxL
Mb 423.03
330)263(15
1 222 =
−=→=+−=′
LLMb 026.0)1423.0)(2423.0(15423.0
=−−=
0.577L 0.384L 0.423L
-0.103L -0.080L
-0.026L
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 147
8.2. Inflence Lines in Members
8.2.1. Moment By the Flexibility Method
– Compatibility Condition:bb
bbbbbb MM
θθ
−=→=θ+θ× ξξ 0
– Betti-Maxwell’s Reciprocal Theorem
ξξ
ξξ θ=θ=→θ−δ=ξ−δ ∫∫ bLb
L
x
L
xb ddxLxdxdx2
2
0
2
0
)2/()(
– Influence Line : bb
b
bb
bb
dM
θ−=
θθ
−= ξξ
Calculation of Deflection
i) Moment Diagram
EIL
bb 38
=θ
L L
ξ Mb = ?? P = 1
ξ P = 1
dxξ , θxξ θbξ
dxb θbb
M = 1
1
2
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 148
ii) Suspended span
baxLEIxw
LxMwEI SS ++−=→−=−=′′
32 3
– Boundary conditions
??224
)0()2
(
00)0(2
=+−→=
=→=
LaEI
LwLw
bw
OS
S
– Deflection of the suspended span
axLEIxwS +−=
3
3
iii) Overhanged span
ecxxL
xEI
wLxMwEI OO +++−=→+−=−=′′ )
23(1)21(
23
– Boundary conditions
026
0)2
(
224)
2()0(
2
2
=++−→=
+−=→=
eLcEILLw
LaEI
LeLww
O
SO
iv) Right span
gfxxL
xEI
wLxMwEI RR +++−−=→−−=−=′′ )
3(1)22( 2
3
– Boundary conditions
EILffLLL
EILw
gw
R
R
320)
3(10)(
00)0(
22
=→=++−→=
=→=
– Deflection
)23(3
1 223 xLLxxLEI
wR +−=
v) Determination of a, c, e
EILc
EILcLL
EIEILL
RO 1217
32)
24(1
32)0()
2( =→=++−→=θ=θ
EILaLa
EIL
EIL
EILee
EIL
EILeLc
EIL
−=→+−=−
−=→=++−→=++−
2242413
24130
2417
60
2622
2222
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 149
vi) Deflection of the left span
– Suspended span: )3(3
13
233
xLxLEI
axLEIxwS +−=+−=
– Overhanged span: )1334128(24
1 3223 LxLLxxLEI
wO +−+−=
Final Influence Line
i) Suspended span
)3(81
38/)3(
31 23
223 xLx
LEILxLx
LEIwM
bb
Sb +=+=
θ−=
LLLLL
LMb 203.06413)
23)
2((
81)
2(
33
2 ==+=
ii) Overhanged span
)1334128(64
1 32232 LxLLxx
LwM
bb
Ob +−+=
θ−=
iii) Right span
)23(81 223
2 xLLxxL
wMbb
Rb +−−=
θ−=
LxLLxxL
Mb 423.00)263(81 22
2 =→=+−−=′ , LLMb 048.0)423.0( −=
0.203L
-0.048L
0.423L
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 150
8.2.2. Influence Line of Shear Force using the Influence Line of Moment
i) 2L
≤ξ
)5(412)3(
4122
012
233
233 xLx
LLxxLx
LLx
LM
VMxLV bbbb −=−+=−=→=−×+×
ii) LL≤ξ≤
2 (Overhanged span)
)1334128(32
120
23223
3 LxLLxxLL
MVMLV b
bbb +−+==→=−×
iii) ξ≤L (Right span)
)23(412
02
2233 xLLxx
LLM
VMLV bbbb +−−==→=−×
EI
L
EI
L
ξ Vb = ?? P = 1
Mb
Vb
χ
P = 1
Mb
0.203L
-0.048L
0.423L
× =≤− )2
(22 LxLx
L
0.406
-0.096
0.423L
-0.594
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 151
8.2.3. Influence line of Shear Force by Müller –Breslau’s Principle
Remove Redunduncy and Apply an Unit Load
bb
xbb d
dV −=
Free Body Digram and Moment Diagram
L
EILdbb 6
4 3
=
L
EI
L
ξ Vb = ?? P = 1
1
1
dxb dbb
L/2
1 1
1
2 1
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 152
Deflection of the Beam
i) Suspended span
baxEIxwxMwEI SS ++−=→−=−=′′
6
3
– Boundary conditions
??8
)0()2
(
00)0(2
=+−→θ=θ
=→=
aEILL
bw
OS
S
– Deflection of the suspended span
axEIxwS +−=
6
3
ii) Overhanged span
ecxxLxEI
wxLMwEI OO +++−=→+−=−=′′ )46
(1)2
( 23
– Boundary conditions
0212
0)2
(
8)
2()0(
2
2
=++−→=
+−=→θ=θ
eLcEI
LLw
aEILcL
O
SO
iii) Right span
gfxxLxEI
wxLMwEI RR +++−−=→−−=−=′′ )26
(1)( 23
– Boundary conditions
EILffLLL
EILw
gw
R
R
30)
26(10)(
00)0(233
=→=++−−→=
=→=
– Deflection
)23(6
1 223 xLLxxEI
wR +−=
iv) Determination of a, c, e
EILc
EILcLL
EIEILL
RO 2417
3)
48(1
3)0()
2(
22222
=→=++−→=θ=θ
EILaa
EIL
EIL
EILee
EIL
EILeLc
EIL
65
82417
48130
4817
120
212332
2333
=→+−=
−=→=++−→=++−
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 153
v) Deflection of the left span
– Suspended span : )5(6
16
233
xLxEI
axEIxwS −−=+−=
– Overhanged span : )1334128(48
1 3223 LxLLxxEI
wO +−+−=
– Right span : ( )xLLxxEI
wR223 23
61
+−=
Final Influence Line
i) Suspended span : )5(41
64/)5(
61 23
3
323 xLx
LEILxLx
EIdwV
bb
Sb −=−=−=
ii) Overhanged span : )1334128(32
1 32233 LxLLxx
LdwV
bb
Ob +−+=−=
iii) Right span : )23(41 223
3 xLLxxLd
wVbb
Rb +−−=−=
0.406
-0.096
0.423L
-0.594
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr