structural analysis istrana.snu.ac.kr/lecture/struct1_2019/notes/19_note(ch1... · 2019-05-03 ·...

Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Structural Analysis I
Spring Semester, 2018
Hae Sung Lee
Dept. of Civil and Environmental Engineering Seoul National University
yδ
yf
zδ zf
xδ
xf
yM yθ
zM zθ xM xθ

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Contents
1. Introduction
2. Reactions & Internal Forces by Free Body Diagra
ms
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

1
Chapter 1
Introduction

2
1.1 Mechanics of Material - Structural Mechanics
Problem
Calculate the reaction force at each support and draw the moment and shear force dia
gram for the two-span beam shown in the figure.
Solution
Equilibrium Equation
qLRRRF cbay 20 =++→=
qLRRLRLRLqLM cbcba 220220 =+→=×−×−×→=
0022
0 =−→=×−×+×+×−→= cacab RRLRLRL
qLL
qLM
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 inde-
terminate structures.
EI EI
q
Ra Rb Rc
L L
q

3
Fundamentals of Differential Equations
Governing Equations
The governing equations of a (engineering) system are usually defined by a system of diffe-
rential equations, which governs behaviors of the given system within a domain. Notice that
the domain does not include boundaries.
11 qwEI =′′′′ for lx

4
1.2 Mechanics of Material
Governing Equation
– Left span
112
13
1
4
1''''
1 24dxcxbxa
EI
qxwqEIw ++++=→=
– Right Span
222
23
2
4
2''''
2 24dxcxbxa
EI
qxwqEIw ++++=→=
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 1
31
4
11111 2402)0( , 0)0( ++=→==′′==
– Right Support
xcxaEI
qxwbwdw 2
32
4
22222 2402)0( , 0)0( ++=→==′′==
x x
y
z
z
y
q
w1 w2

5
– Center Support
==
−==→
+=+
−−−=++
=++=
=++=
EI
qLcc
EI
qLaa
LaEI
qLLa
EI
qL
cLaEI
qLcLa
EI
qL
LcLaEI
qLLw
LcLaEI
qLLw
48
48
3
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 xLLxxEI
qww +−=≡
83
, 8
32 11
211
qLqxwEIVx
qLx
qwEIM +−=′′′−=+−=′′−=
Moment Diagram
Shear Diagram
Reactions
0.375qL
L83
+
-
+
0.625qL
-
0.375qL 0.375qL 1.25qL
0.125qL2
0.070qL2
L8
3
+
-
+

6
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 (compatibility condition)
δ0+ δR=0
1.3.2 Calculation of δ0
Bending Moment
qLxqx
M +−=2
2
q
q
δ0
δR
Rb
q
qL2/2

7
Governing Equation
baxqLxqx
EIwMwEI ++−=→−=′′624
34
00
Boundary (support) Conditions
– Left Support : 0)0(0 =w
– Right Support : 0)2(0 =Lw
Determination of Integration Constant
– Left Support
0 0)0(0 =→= bEIw
– Right Support
EI
LqaLa
LqLqLEIw
24
)2(0)2(
6
)2(
24
)2()2(
334
0 =→=+−=
Deflection
))2()2(2(24
3340 LxLxxEI
qw +−=
EI
LqLLLLL
EI
qLw
384)2(5
))2()2(2(24
)(4
33400 =+−==δ
1.3.3 Calculation of δR
Bending Moment
22,
22
21
1
LRxRM
xRM bbb −=−=
Governing Equation
+++−=
++=→
−=′′−=′′
222
22
32
2
111
31
1
22
11
412
12
bxaLxRxR
EIw
bxaxR
EIw
MwEI
MwEI
bbR
bR
R
R
1x 2x
RbL/2
Rb

8
Boundary (support and mid-span) Conditions
– Left Support & Right Support
0)( , 0)0( 21 == Lww RR
– Mid-span
)( )0()()( , )0()( 221121 LwLwLwLw RRRRRR θ=′=′=θ=
Determination of Integration Constant
– Left Support
0 0)0( 11 =→= bwR
– Right Support & Mid-span
−=
=
−=
→
′==+=′
==+=
=+++−=
6
04
(0)4
)(
(0)12
)(
0412
)(
3
2
2
2
1
221
2
1
221
3
1
22
33
2
LRb
a
LRa
wEIaaLR
LwEI
EIwbLaLR
LEIw
bLaLRLR
LEIw
b
b
Rb
R
Rb
R
bbR
Deflection
)23(12
)3(12
322
322
123
11
LLxxEI
Rw
xLxEI
Rw
bR
bR
+−−=
−=
EI
RLwLw bRRR 48
)2()0()(
3
21 −===δ
Compatibility Condition
δ0+ δR=0 → 048
)2(
384
)2(5 34 =−EI
RL
EI
Lq b → qLRb 810=

9
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
EI
Lqdx
EI
MML R384
)2(5 42
0
00 −== δ , EI
LRdx
EI
MM bL
RRR 48
)2( 32
0
== δ
Solution
δ0+ δR=0 → 048
)2(384
)2(5 34 =+−EI
RL
EI
Lq b → qLRb 810=
RbL/2
Rb
q
q
qL2/2

10
– Moment
– Shear
+
-
+
-
+
+ -
+
-
=
0.070qL2
5qL2/8
Rb
+
=
0.125qL2 L
8
3
+
-
+
qL2/2

11
1.5 Supports
– Supports are used to maintain the position of the structure by constraining free mo-
tions at some discrete points.
– If you want to constrain a free motion of a structure at any part of the structure, some
kind of forces should be applied.
– Since a support prevent free motion of a structure, a force called as a reaction force is
developed in the direction of a constraints.
– The support conditions provide boundary conditions for the governing equation(s).
– The reaction forces are unknown, and should be calculated through the structural
analysis.
– The most popular support types are a fixed support, a hinge support and a roller sup-
port.
δ0δ

12
Fixed Support(고정단)
– Both the horizontal and vertical motion and rotational motion are prevented.
– Reaction forces are developed in three directions.

13
Hinged Support(회전단)
– The horizontal and vertical motion are prevented, but the rotational motion is allowed.
– The horizontal and vertical reaction forces are developed in all three directions
Roller Support(이동단)
– Only the vertical motion is prevented, and the horizontal and the rotational motion is
allowed.
– A reaction force is developed only in the vertical direction.

14
1.6 Two-Dimensional Idealization of a Structure
Real 3-D Structures
– 3 force components and 3 moment components
– 3 displacement components and 3 rotational components
Main Structures : A part of a structure designed mainly to resist external loads.
– Beams, Truss, Frame, Arch…
Secondary Structure : A part of a structure designed to transfer external loads to
main structures or increase stability and/or stiffness of a structure
– Stringer : Small beam that runs parallel to a main structure to support the floor
system.
– Cross beam : Small beam that runs perpendicular to the main structure to support
stringers.
– Bracing (tie) : member that connects two or more main structures to increase mainly
lateral stability of the main structures.
xδ xf
ym yθ zδ
x
y
z
yδ yf
zδ zf
xδ xf ym yθ
zm zθ xm xθ

15
Load transfer path
External load → Slab → Stringer → Cross Beam → Main Structure → Support →
Foundation
– External loads applied on a slab are supported by stringers, which transfer the carried
loads to cross beams. The cross beams, in turn, transfer the carried-over loads from
the stringers to main structures.
주 구조물 (Main Structure)
가로 보 (Cross Beam)
세로 보 (Stringer )
Cross Bracing (Wind Bracing)
지 점 (Support)

16
Two-dimensional Idealization of a Structures
All members are idealized as one-dimensional members, which are represented by
lines passing the centroids of the cross sections, because dimensions in cross-
sectional directions are very small compared to their lengths (usually less than 1/20).
Suppose we somehow know portions of the external loads transferred to each main
structure.
Suppose all members in each main structure lie in one plane, and external loads car-
ried-over from substructures are applied on the same planes.
Then, all structural responses are developed in the same plane, and main load resist-
ing actions occur in the same plane.
If this the case, each main structure in the same plane can be separated from the
whole structure by neglecting three-dimensional effects, and analyzed in the plane
where each main structure is defined for the simplicity of a structural analysis.
Two-dimensional structures are often referred to as plane structures.
However, fundamentals employed in the 2-D analysis of structures are exactly the
same as those in the 3-D analysis.

17
Plane truss idealization
– Structure composed of a number of bars carrying axial forces that lie in one plane.
– Vertical and horizontal force
– Vertical and horizontal displacement
절점
xf xδ
yδ yf

18
Plane beam idealization
– Structure composed of a number of beams carrying bending moments and shear
forces that lies in one plane.
– Shear force and Moment on z-axis
– Vertical displacement and rotational angle with respect to z-axis
11 , wV 22 , wV
33 , wV
11 , θM 22 , θM 33 , θM

19
Plane Frame idealization
– Structure composed of a number of beams carrying bending moments, shear forces
and axial forces that lie in one plane.
– Vertical, horizontal force and moment w.r.t. z-axis
– Vertical, horizontal displacement rotational angle w.r.t. z-axis
xδ xf
yf yδ zM zθ

20
1.7 Stability of Structures
Internal Stability (내적 안정)
A structure maintains its shape under the actions of all possible loads within a certain limit.
External Stability (외적 안정)
A structure maintains its position under the actions of all possible loads within a certain limit.

21

22
Chapter 2
Reactions & Internal Forces
by Free Body Diagrams

23
2.1 Free Body Diagram
Any portion of a structure can be isolated by passing any desired section through structure.
A free body sketch is a diagram drawn showing this portion acted upon by the external loads
and reactions, together with any forces that may be act on the faces of the members cut by the
isolating section. Or, a diagram drawn by removing some part(s) of a structure and replacing
the removed part(s) with unknown forces.

24
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 -

25
2.2 Reactions
Beams
→
+=−→=
=−+→=
) (clockwise 00
00
LRPaM
PRRF
BA
BAVP
L
aRB = , PL
bRA =
→
+=−+→=
=+→=
) (clockwise 0)(0
0
LRaLPM
PRRF
BA
BAVP
L
aRB )1( += , PL
aRA −=
Truss
PHPHF AAH −=→=+→= 00 ,
00 =−+→= PRRF BAV
0)3(0 =−+→= aRPaPaM BA (Clockwise +)
PRA 31= , PRB 3
2=
P a b
L
P
RAL
RBL
b a
P a
L
P a
L RAL
RBL
RAL
RB P
P
HAL
P
P
[email protected]
a

26
Frame
+=
−=→
=−−=
=−−=
=+=
=−→=−+=−+=
42
2
0422
02
082
24
2 02
qLHR
HR
LqLLH
LRM
LHL
RM
HHF
qLHH
qLH
qLH
qLRRF
AA
BB
AALh
BBRh
BAH
BABABAV
What if there is no internal hinge ?
042
)(2
)(
0422
02
0
02
=−+−−→
=−−+−=
=−−=
=+=
=−+=
LqLLHH
LRR
LqLLH
LRMM
LHL
RMM
HHF
qLRRF
ABBA
AALh
BBRh
BAH
BAV
HB
RA RB
HA
L
L
h
q
→−=
=
16
16qL
H
qLH
B
A
8
8
3
qLR
qLR
B
A
=
=
HB
RA RB
HA
q
M M

27
2.3 Internal Forces in Framed Structures
Axial Force
Shear Force
Bending Moment
Torsion
+
+
+
+

28
2.4 Internal Forces in Simple Beams
Reactions
q
RA=qL/2 Rb= qL/
q
RA=qL/2 RB= qL/2
RA RB

29
Free Body Diagram for Shear and Moment
qxqL
qxRVVqxRF AxxAV −=−=→=−−= 20
22
02
2qxx
qLMM
xqxxRM xxAx −=→=−−=
Differential Equation
baxxq
Mqdx
Md ++−=→−= 22
2
2
Boundary Conditions: 2222
,00)()0( xq
xqL
MqL
abLMM −=→==→==
Shear Force and Moment Diagrams
Deflected Shape
Remember 2
2
dx
wdEIM −= !!!
RA
x
RB
Mx
Vx
+
qL/2
qL2/8

30
2.5 Gerber Systems
A structural system that an unstable structural component is supported by a stable structure.
Sometimes there may be several unstable structures are supported by a stable structure, and
vice versa. An external load applied to an unstable structure is transferred to a stable struc-
ture, but an external load applied to a stable structure is not transferred to an unstable structure.
In the later case, a unstable structure is just deflected as a rigid body without deformation.

31

32
2.5.1 Internal Forces in a Gerber Beam - I
Reactions
PRL
PL
RM HHC 3
20
24
30 =→=×−×→=
PRPRRF CCHv 3
100 =→=−+→=
PRL
RLRM BHBA 6
50
4
50 =→=×+×−→=
PRPRRF ABAv 6
10
3
20 −=→=−+→=
L/4
P
RH
RA RB
RC
P
P/6 5P/6 P/3
P

33
Shear Force
i) Lx ≤≤0
ii) LxL2
3≤≤
iii) LxL 22
3 ≤≤
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= Px6
1−

34
ii) LxL2
3≤≤ , Lxx −=′
iii) LxL 22
3 ≤≤ xLx −=′ 2
Deflected Shape
qdx
wdEI =
4
4
or 2
2
dx
wdEIM −=
P/6 5P/6
2P/3
Mx=
xPPL ′+− 2
P/6 5P/6
P
P/3 P/3
Mx= xP ′3
1
+
-
PL/6
PL/6

35
2.5.2 Internal Forces in a Gerber Beam - II
Free Body diagram
Shear
q
2
ql
2
ql
2
ql
2
ql
2
2ql
2
2ql
2
ql
2
ql
+
L
q
L L
2
ql
2
ql
x
qxql
Vx −= 2 2
ql

36
Moment
Deflected Shape
8
2ql
2
2ql 2
2ql
+
xqlql
Mxqlql
M xx 220
22
22
+−=→=−+
2
ql
2
2ql
2
ql
x
2
ql
220
2
2qxx
qLMM
xqxxR xxA −=→=++−

37
2.6 Truss (트러스)
Structures composed of a number of straight bars carrying axial forces.

38
Assumption
1. All joints are hinges.
2. All members are straight.
3. Small deformatiom
4. The external loads are applied only at joints.
The 1st assumption seems to be unreasonable, but it is very reasonable assumption !!!
Because the length of each member is short compared to the total length of a truss, the mo-
ment induced by rigid joints is negligibly small in an engineering sense.
Characteristics of truss
– By the 2nd , 3rd and 4th assumptions
02
2
2
2
=→−=dx
Mdq
dx
Md→ baxM +=
– By the 1st assumption
0 , 000)()0( ≡≡→==→== VMbaLMM
– 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.
– The axial force is constant along a member by assumption 1.
– There is only one unknown per member.
– As the equilibirium of each member is always satisfied, the equilibrium conditions at
joints are to be considered.
– Determinancy
– The number of unknowns : RM NN +
– The number of Equilibrium Equations : JN2
– 02 =−+ JRM NNN : Determinant, 02 >−+ JRM NNN : Indeterminant

39
2.6.1 Internal Forces in Howe Truss
At U1 and U3
At L1
2 ,
22
022
2
02
2
23
3
23 PFPF
PF
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
9+3-2×6=0: determinant truss
L2
7
9 1
4
2
3
U1 U2 U3
6
5
L1
P/2 P/2
L3
8
P
[email protected]
a

40
At U2
=+−
−==→
=−−−
=++−−
02
1
2
12
2
02
2
2
2
02
2
2
237
753
8734
PPP
PFF
FFF
FFFF
At L3
Because we already have 3 equilibrium equations to calculate the reactions
Axial Force Diagram
2
P
2
P
F5
F4 F8
F3 F7
2
P 0
P/2
P/2
P/2 P/2
P 0
0 0
0
P P/2 P/2
Tension Compression

41
Equilvalent Beam Action
Axial forces in members generate bending moments and shear forces just like in a beam !!!
Deflected Shape
2.6.2 Internal Forces in Warren Truss
V=P/2
P/2
1 P
3P
32P
2 3
4
5
6
7
8
9
10
11
L1 L2 L3
U1 U2 U3
L4
x
PL/4 M=Px/2
[email protected]
l

42
At L1
PFPFP
F
FF
32
, 3
22
03
222
022
12
2
12
=−=→
=+
=+
At U1
PFPFF
FF
FFF
34
,232
022
22
02
2
2
2
423
32
432
−==−=→
=+
=++−
At L2
PFPF
FFFF
PFF==→
=+−+
=−+65
5361
35
,32
022
22
-
022
22
F2
F1
2P/3
F4
F2 F3
F3
F1 F6
P
F5

43
At U2
PFPF
FF
FFFF
32
,32
022
22
02
2
2
2
87
75
7584
−=−=→
=−−
=+−+−
At L3
3,
32
022
22
022
22
109
97610
97 PFPF
FFFF
FF==→
=+−−
=+
At U3
PF
FF
FFF
32
022
22
022
22
11
119
8119
−=→
=−−
=−+−
F8
F5 F7
F4
F7
F6 F10
F9
F11
F8
F9

44
At L1
OK
Deflected Shape
Equivalent Beam Action
1 2 3 4 5 6
1. PxM32=
2. PxPlM32
32 +=
3. PxPlxlPPlM31
34
)(31 −=−+=
4. PxPlxlPPlM31
)(31
32 −=−+=
5. PxPlxlPPlM31
32
)(31
31 −=−+=
6. PxPlxlPM31
31
)(31 −=−=
P/3
P/3
P32
P
Pl3
4
PxPlM3
2
3
2 += 2P/3
2P/3
4P/3=2P/3+2P/3
2P/3 x
x
2P/3
PxM3
2=

45
2.6.3 Method of Sections
PFL
FLPM L 34
023
2442 −=→=×+×=
PFL
FL
PLPMU =→=×−×−×= 662 02223
32
PFFPPFV 32
022
32
55 =→=×+−=
P
3P
P32
1
2 3
4
5
6
7
8
9
10
11
L1 L2 L3
U1
U
U2 U3
L4
Cut out
P
P32
F5
F4
F6

46
2.7 프레임 (Frame)

47
2.7.1 Internal Forces in a Frame
Reactions
Freebody Diagram
H
qH
L
L
qH
2
2
q
L
qH
2
2
2
2qH
qH
L
qH
2
2
L
qH
2
2
2
2qH
L
qH
2
2
LqH
2
2
L
qH
2
2
L
qH
2
2

48
Axial, Shear and Moment diagram
Deflected Shape
L
qH
2
2
+ - Axial
L
qH
2
2
qH
-
+ Shear
L
qH
2
2
2
2qH
+
Moment
+

49
2.7.2 Internal Forces in a 3-hinged Frame
Reactions (+:Clockwise for mement)
02
=+
=+
BA
BA
HH
qLRR
= 0RhM : BBBB HRLHLR 202 −=→=−−
= 0LhM : 420422qL
HRLqL
LHL
R AAAA +=→=−−
082
24
2
=+
=−→=−+
BA
BABA
HH
qLHH
qLH
qLH
16
16qL
H
qLH
B
A
−=
= →
8
83
qLR
qLR
B
A
=
=
L
L
HA HB
RA RB
h
q

50
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
16qL
- +
- 8
qL
+
83qL
Shear
83qL
8qL
16qL
- -
-
Axial

51
283
16
028
316
22
2
xqqLx
qLM
xqxqLx
qLM
−+−=
=−+−−
Deflected Shape
- -
-
Moment
- -
2
16qL 2
16qL
83qL
16
2qL
M
V

52
2.8 Arches

53
2.8.1 Three Hinged Arch
Arch Curve : )( 22
2xl
l
hy −=
Reactions
2
2P
R
PR
B
A
=
= ,
h
PlH
h
PlH
B
A
2
2
=
=
Freebody Diagram
h
PlH
PV
2
2
=
=
0)( =++−− xLRyHM AA
)(2
)(2
)(2
)(22
2
222
xlxl
P
xlP
xll
h
h
Pl
xlP
yh
PlM
+=
++−−=
++−=
8maxPl
M −=
HA
RA
HB
RB
P
h
2l
x
y
HA
RA
V
H
M
y
l+x

54
Axial force and Shear Force
θθθ−θ
=
→
θ+θ=θ−θ=
S
A
V
H
SAV
SAH
cossin
sincos
cossin
sincos
θθ−θθ
=
→
V
H
S
A
cossin
sincos
224224
2
2
4
2sin ,
4cos
2tan
xhl
hx
xhl
l
l
hxy
+
−=θ+
=θ
−=′=θ
)2
(4
cos2
sin2
)2
(4
sin2
cos2
2
224
23
224
llx
xhl
PP
h
PlS
h
xhl
xhl
PP
h
PlA
−−+
=θ−θ=
−
+−=θ−θ−=
Deflected Shape
V
H
A S θ

55
2.8.2 Zero Moment Arch I
Reactions
2
2P
R
PR
B
A
=
= ,
h
PlH
h
PlH
B
A
2
2
=
=
Freebody Diagram
h
PlH
PV
2
2
=
=
0)( =++−− xLRyHM AA
)(
0)(22
xll
hy
xlP
yh
PlM
+=
=++−=
HA
RA
HB
RB
P
h
2l
x
y
H
HA
RA
V
M
y
l+x
x
y

56
2.8.3 Zero Moment Arch II
Reactions
qlR
qlR
B
A
== ,
h
qlH
h
qlH
B
A
2
22
2
=
=
Freebody Diagram
02
)()()( =++−++−− xlxlqxlRyHM AA
)(
02
)()()(
2
222
2
xll
hy
xlxlqxlqly
h
qlM
−=
=++−++−=
HA
RA
HB
RB
q
h
2l
x
y
HA
RA
V
H
M
y
l+x
q

57
Chapter 3
Principle of Virtual Work
The principle of virtual work is the most important subject in the area of the structural analysis !!!!

58
3.1 Beam Problems
3.1.1 Governing Equations
Equilibrium for vertical force
qdx
dVqdxVdVV −=→=+−+ 0)(
Equilibrium for moment
Vdx
dMdxqdxVdxMdMM =→=+−−+ 0
2)(
Elimination of shear force
qdx
Md −=2
2
Strain-displacement relation
A plane section before deformation remains a plane after deformation.
A section perpendicular to the neutral axis before deformation maintain the right
angle to the neutral axis after deformation
ydx
wd2
2
−=ε
Stress-strain relation (Hooke’s law)
ydx
wdEE
2
2
−=ε=σ
Definition of Moment
2
22
2
2
dx
wdEIdAy
dx
wdEydAEydAM
AAA
−=−=ε=σ=
Beam Equation
qdx
wdEI =
4
4
M M+dM
V V+dV
q

59
3.1.2 Limit Process
Equilibrium for vertical force
qdx
dVq
x
Vq
x
VxqVVV
x−=→=+
∆∆→=+
∆∆→=∆+−∆+
→∆0)(lim00)(
0
Equilibrium for moment
0))()(1
(lim0)()()(0
00
=ξξ−∆ξ+∆
+−∆
∆→=ξξ−∆ξ++∆−−∆+ ∆
→∆
∆
dxxqx
Vx
MdxxqxVMMM
x
x
x
00)(),()0,(),(
lim)(1
lim)(lim
)(),(
00
00
00
=×−=ξ
ξ−=∆
−∆−=ξξξ+∆
−ξξ+
ξξξ+=ξ
=ξ→∆
∆
→∆
∆
→∆
xqd
xdQ
x
xQxxQdxq
xdxq
dxqxQ
x
x
x
x
x
3.1.3 Modelling of Concentrate loads - Dirac delta functions
0lim→ε
= = )( ξ−δ x
122
1lim)0
2
10(lim)(
00
00
=εε
=+ε
+=ξ−δ→ε
ε+ξ
ε+ξ
ε−ξ
ε−ξ
→ε ll
dxdxdxdxx
)()(2
)()(lim)(
2
1lim
)0)(2
1)(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
fFFF
dxxf
dxxfdxxfdxxfdxxxfll
ξ
∞
ξ
ε21
2ε
ξ
M M+∆M
V V+∆V
q
∆x

60
3.1.4 Principle of Virtual Work (Beam)
The real(actual) load case A virtual(imaginary) load case
For the real load case
02
2
=+ qdx
Md,
2
2
dx
wdEIM −= for lx

61
Principle of virtual work
dxEI
MMdxM
dx
wd ll
−=00
2
2
dxqwdxEI
MM ll =00
→ extWW δ=δ int
The internal virtual work is equal to the external virtual work if a beam satisfy the
equilibrium for the real and virtual load.
Equilibrium equation for load case q
0)(0
2
2
=+ dxqdxMd
wl
Virtual work expression
dxqwdxEI
MM ll
=00
Betti-Maxwell’s Reciprocal Theorem
dxEI
MMdx
EI
MM ll
=00
→ dxqwdxqwll
=00
Calculation of displacement for the load case q
dxqwdxEI
MM 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
0
0
xwdxxxwdxEI
MM ll =−δ= → dxEIMM
xwl
=0
0 )(

62
3.1.5 Example
A simple beam subject to an uniform load
– Moment of load case q
– Moment of load case q
– Deflection at the center of the span
EI
ql
EI
qllqlql
EI
dxxq
xql
EIdxx
qx
qlx
EI
dxxq
xqlxl
EIdxx
qx
qlx
EIdx
EI
MMlw
ll
l
l
ll
4443
2/
0
322/
0
2
2/
22/
0
2
0
384
5)
256
1
96
1(
2))
2(
4
1
4)
2(
3
1
4(
2
)44
(2
)22
(2
2
)22
)(22
(1
)22
(2
1)
2(
=−=−=
−=−=
−−+−==
or from the integration table,
EI
qlqll
EI
lMM
l
ab
EI
ldx
EI
MMlw
l
3845
84)
41
1(3
)1(3
)2
(42
3120
=+=+==
l/4
1
q
ql2/8

63
Values of Product Integrals L
LU dxMM0

64
3.1.6 Conservation of Energy
Equilibrium and Conservation of Energy
– Equilibrium Equation
qdx
wdEI =
4
4
– External work
int
ll
lll
lll
ll
ext
WdxEI
Mdx
dx
wdEI
dx
wd
MwVdxdx
wdEI
dx
wd
dx
wdEI
dx
dw
dx
wdwEIdx
dx
wdEI
dx
wd
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
2
1
2
1
][2
1
][2
1
2
1
2
1
Conservation of Energy in each load case
dxwqdxEI
M ll
=00
2
21
21
, dxqwdxEI
M ll
=00
2
21
21
Two load cases are applied simultaneously.
dxqwqwdxEI
MM
dxwqqwqwqwdxEI
Mdx
EI
MMdx
EI
M
dxqqwwdxEI
MM
ll
llll
ll
+=
+++=++
++=+
00
00
2
00
2
00
2
)(2
1
)((2
1
2
1
2
1
))((2
1)(
2
1
w
q
w
q

65
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
dxqwdxqwdxqwqwdxEI
MM llll
==+=0000
)(2
1
3.1.7 General Conservation and Equilibrium
Conservation in General
=+⋅−S V
fdVdS 0nv
v n
dS
w
q
w
q
ww +
w
q
w
q
ww +qq + qq +

66
– By divergence theorem,
⋅∇−=⋅−VS
dVdS vnv where ),,(),,(321 xxxzyx ∂
∂∂∂
∂∂=
∂∂
∂∂
∂∂=∇
=+⋅−∇=+⋅∇−=+⋅−VVS VV
dVffdVdVfdVdS 0)( vvnv
– Since the integral equation should hold for all systems,
0=+⋅∇− fv
Potential Problems
– The vector field of a system is defined by a gradient of a scalar function referred to as
a potential function
Φ∇⋅−= kv , ),,(zyx ∂Φ∂
∂Φ∂
∂Φ∂=Φ∇
– The famous Laplace equation for a conservative system.
0)( =+Φ∇⋅⋅∇=+⋅∇− ff kv
– If the system properties are homogeneous and isotropic, Φ∇−= kv
0)( 2 =+Φ∇=+Φ∇⋅∇=+⋅∇− fkfkfv or 02
2
2
2
2
2
=+∂
Φ∂+∂
Φ∂+∂
Φ∂f
zyx
3.1.8 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 =
⋅=σ=3
1jijiji nT nσ ,
=
σσσσσσσσσ
=
3
2
1
333231
232221
131211
σσσ
σ ,
=
3
2
1
n
n
n
n
TT=
E, ν
uu = Su
St
V
x
y
z
bx
by
bz

67
– Divergence Theorem
0)( =+⋅∇=
+⋅∇=+⋅=+
V
ii
V V
ii
S V
ii
S V
ii
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 bx
bzyx
bσ for i = 1,2,3 or
0
0
0
3333231
2232221
1131211
=+∂σ∂
+∂σ∂
+∂σ∂
=+∂σ∂
+∂σ∂+
∂σ∂
=+∂σ∂
+∂σ∂+
∂σ∂
bzyx
bzyx
bzyx
Moment Equilibrium : 0= iM for i=1, 2, 3 or 0=M
0=+×+× VVS
dVdVdS mfxvx
211231133223 , , σ=σσ=σσ=σ
– In case elasticity problems are potential problems
ii
i uC ∇⋅=σ
– However, to maintain symmetry condition of stress,
uC ∇= :σ or = = ∂
∂=σ3
1
3
1k l l
kijklij u
uC
– Equilibrium equation in terms of the potential functions
0):( =+∇⋅∇ buC
– What is σ, and why is σ related to a potential function in such a way?
out of scope of this class !

68
3.1.9 Displacement on boundaries
Virtual Work Expression
llll
MVwdxqwdxEI
MM00
00
θ−+=
llll
MVwdxqwdxEI
MM00
00
θ−+=
)0()0()()()0()0()()(0
0000
MlMlVwlVlwdxqwMVwdxqwdxEI
MM lllll θ+θ−−+=θ−+=
By coinciding the positive direction of forces and displacement
)0()0()()()0()0()()(00
MlMlVwlVlwdxqwdxEI
MM ll θ+θ+++=
Deflection of a cantilever beam subject to an end load
0)0( , 0)0( , 0)( , 1)( =θ=== wlMlV
q (real) system q (virtual) system
)0()0()()()0()0()()(00
MlMlVwlVlwdxqwdxEI
MM ll θ+θ+++=
EI
PlPll
EI
lMM
EI
ldx
EI
MMlw
l
3))((
33)(
3
31
0
=−−===
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
)0()0()()()0()0()()(00
MlMlVwlVlwdxqwdxEI
MM ll θ+θ+++=
EI
PlPl
EI
lMM
EI
ldx
EI
MMl
l
2)(1
22)(
2
31
0
−=−××===θ
P
-Pl
1
-l
P
-Pl
1

69
Rotation in the a body (or a structure)
– Modeling of a unit moment applied at x0
)0()0()()()0()0()()(00
MlMlVwlVlwdxqwdxEI
MM ll θ+θ+++=
)()]2
()2
([1
lim
))]2
((1
))2
((1
[lim
0000
0
000
0
0
xdx
dwxwxw
dxxxxxwdxEI
MM
xx
ll
θ−=−=ε+−ε−ε
=
ε+−δε
−ε−−δε
=
=→ε
→ε
– by coinciding the positive direction of the rotational angle with that of the applied
moment.
dxEI
MMx
l
=θ0
0 )(
ε 1/ε 1/ε
x0

70
3.2 Principle of Virtual Work in General
3.2.1 3-Dimensional Elastic Body
Rigid Body
Rigid body: A body that does not deform at all. Displacement is constant over the body.
Deformation of a body is caused by the variation of displacements in a 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 bTwbwTw
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 !!!
q-Force System
q -Force System
w
T
T
dVdSij V
ijij
S
σε=⋅ Tw

71
⋅+⋅=δS V
ext dVdSW bwTw
=δV
dVW σε :int
= == =
= == ==
∂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 V
ijij
i j S
jijii S j
jijii S
ii
S
dVx
wx
wdVw
x
dSnwdSnwdSTwdSTw
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
)(2
1
)(2
1
)(2
1
)(
)(
WdVdVx
w
x
w
dVx
wdV
x
w
dVx
wdV
x
wdV
x
w
dVbx
wdVx
w
dVbwdVx
wx
wW
i j V
ijiji j V
iji
j
j
i
i j V
iji
j
i j V
ijj
i
j i V
jii
j
i j V
ijj
i
i j V
ijj
i
i V ji
j
iji
i j V
ijj
i
i V
iii j V j
ijiij
j
iext
δ=σε=σ∂∂
+∂∂=
σ∂∂
+σ∂∂=
σ∂∂
+σ∂∂=σ
∂∂=
+∂σ∂
+σ∂∂=
+∂σ∂
+σ∂∂=δ
= == =
= == =
= == == =
= == =
== =
=⋅+⋅S VV
dVdVdS σε :bwTw

72
3.2.2 General Framed Structures
)( τγ+σε=τγ+σε=σ⋅εe VVVVij V
ijij dVdVdVdVdVee
Internal virtual work by normal stress – bending moment
=−−=
==
==−−=σε
ee
ee
e
e
eeee
ll
ll
A
l
AVVV
dxEI
MMdx
EI
MEI
EI
M
dxdx
wdEI
dx
wddx
dx
wddAEy
dx
wd
dAdxydx
wd
dx
wdEdVy
dx
wd
dx
wdEdVy
dx
wdEy
dx
wddV
00
02
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
FFdx
A
FdA
EA
FdV
A
F
EA
FdV
00
)(
Internal virtual work by shear stress
QyIb
V
)(=τ and Q
yGIb
V
)(=γ where =
a
y
ydAQ
===τγee
eee
ls
l
AVV
VdxVGA
fdAdx
ybI
Q
G
VVQdV
yIb
VQ
yGIb
VdV
0022
2
)()()(
Total displacement
++=e
l ll
s
e ee
dxEA
FFdx
GA
VVfdx
EI
MMxw
0 00
0 )()(

73
3.2.3 Effect of Shear Deformation
For simple beam with a uniform load case
EI
qllwM 384
5)
2(
4
=
Shear Effect
▬ Shear force of load case q
▬ Shear force of load case q
▬ Deflection by shear force
GA
qlfVV
l
GA
fVdxV
GA
fVdxV
GA
fw ss
ls
ls
S
e
822
122 22/
00
====
GA
EI
l
f
EIql
GAqlf
w
w ss
M
s
40384
384/58/
24
2
==
for a rectangle section of bh× with steel
2
2
2
3
5.21240
6.238456
l
h
bhl
bh
w
w
M
s =×
××=
For small h/l, the effect of shear deformation can be neglected.
1/2
ql/2

74
3.3 Truss Problems
3.3.1 Principle of Virtual Work
From General principle
=++=
α+=α=⋅
ii
ii
e
l ll
s
iii
S
lEA
FFdx
EA
FFdx
GA
VVfdx
EI
MM
vuwdS
e ee
0 00
22
)(
coscoswq
From equilibrium equation
0 , 0)(
1
)(
1
=+−=+− ==
iim
j
ij
iim
j
ij YVXH for njni ,,1L= (
ij
ij
ij
ij
ij
ij FVFH θ=θ= sin , cos )
where m(i), ijH and ijV 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−
njn
i
iiiinjn
i
im
j
ij
ij
iim
j
ij
ij
i
njn
i
iiim
j
ij
ij
iiim
j
ij
ij
vYuXFvFu
vYFuXF
11
)(
1
)(
1
1
)(
1
)(
1
) ()sin cos(
0))sin( )cos((
Each member has two ends, and each end should be connected to two different joints.
Therefore adding all member end forces in a joint-wise fashion is equivalent to adding all
member end forces in a member-wise fashion because the member end forces of a
member
iY
iX
iF1−
ijF−
i imF )(−
iY
iX
pqF−
ijF ijF−
pqF
i-th joint
p-th joint
k-th member
1
2

75
==
== ==
+=−θ+−θ
+=θ+θ
njn
i
iiiinmb
ikkkkkkkk
njn
i
iiiinjn
i
im
j
ij
ij
iim
j
ij
ij
i
vYuXvvFuuF
vYuXFvFu
11
1212
11
)(
1
)(
1
) ())(sin )(cos(
) ()sin cos(
==
====
=+
==∆=−θ+−θ
nmb
k k
kkknjn
i
iiii
nmb
k k
kkk
k
kknmb
kk
nmb
kkk
nmb
kkkkkkkk
EA
lFFvYuX
EA
lFF
EA
lFFlFvvuuF
11
1111
1212
) (
))(sin )((cos
Since α represnts the angle between the applied unit load and the displacement vector, αcosiu are the displacement of the joint i in the direction of the applied unit load.
For vertical displacement For Horizontal displacement
iu αcosiu
kkk vv θ− sin)(12
kk uu − cos)(12
kθ
kθ
)( 12 kk uu −
)( 12 kk vv −By applying Betti-Maxwell reciprocal
theorem,
==
=+nmb
i k
kkknjn
i
iiii
EA
lFFvYuX
11
) (
The displacement of a joint i in a truss is
obtained by applying a unit load at a joint i
in an arbitrary direction.
=
=α=
α=+nmb
k k
kkki
iiiii
EA
lFF
vYuX
1
cos
cos
u
uX

76
3.3.2 Example
Real System Virtual System
Table for calculation of the deflection of a truss
Member EA
l 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
EA
l
EA
L215)260130( =+=δ
1
2
3
4
5
6
7
8
9
10
11
12
13
20 20 20

77
3.3.3 Conservation of Energy (Truss)
Equilibrium and Conservation of Energy
▬ Equilibrium Equation
0 , 0)(
1
)(
1
=+−=+− ==
iim
j
ij
iim
j
ij YVXH for njni ,,1L=
▬ External work
= ====
+=⋅∆=+=njn
i
iim
j
ij
iim
j
ij
njn
iii
njn
i
iiiiext vVuHvYuXW
1
)(
1
)(
111
) (21
21
)(21
P
int
nmb
k k
kkknmb
kkk
nmb
kkkkkkkkk
njn
i
im
j
ij
ij
iim
j
ij
ij
i
njn
i
iim
j
ij
ij
iim
j
ij
ij
njn
i
iim
j
ij
iim
j
ijext
WEA
lFFlF
vvFuuF
FvFu
vFuFvVuHW
==∆=
−θ+−θ=
θ+θ=
θ+θ=+=
==
=
= ==
= === ==
11
1
1212
1
)(
1
)(
1
1
)(
1
)(
11
)(
1
)(
1
2
1
2
1
))(sin )(cos(2
1
)sin cos(2
1
)sin cos(2
1) (
2
1
intext WW =
Conservation of Energy in each load case
==
⋅∆=njn
iii
nmb
k k
kk
EA
lF
11
2
21
21
P , ==
⋅∆=njn
iii
nmb
k k
kk
EA
lF
11
2
2
1
2
1P
Two load cases are applied simultaneously
====
⋅∆+⋅∆=→+⋅∆+∆=+njn
iiiii
nmb
k k
kkknjn
iiiii
nmb
k k
kkk
EA
lFF
EA
lFF
1111
2
)(2
1)()(
2
1)(
2
1PPPP
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
k k
kkk
EA
lFF
111
PP
===
δ=δ=njn
kkk
njn
kkk
nmb
k k
kkk lFlFEA
lFF
111

78
3.4 Frame Problems
++=∆e
l ll
s
e ee
dxEA
FFdx
GA
VVfdx
EI
MM
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
EI
PllPlllPll
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

79
+=δ
2/
00
2 lls
S dxdxGA
f
GA
Plf)
PlPl(
GA
fdx
GA
VVf ss
V
sS 8
3
2
1
224
1
4
2 =××+××== δ
+=
2/
00
2 lls
A dxdxEAδ
EA
PlPlPl
EAdx
EA
AA
V
A 16
9)
4
1
422
1
2(
2 =××+××==δ
))(75.0)(56.11(16
)961(16
223
22
3
l
h
l
h
EI
Pl
EAl
EI
GAl
EIf
EI
Pl sASM
++=
++=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
dxEI
MM
0
×
P/4 1/4
×
P/2 1/2
×
P/4 1/4
×
P/2 1/2

80

81
Chapter 4
Analysis of Statically Indeterminate Beams
Rxθ L
xθ
1

82
4.1 Propped Cantilever Beam
Equilibrium equation
02
0
=−+
=+−−
lRl
qlM
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
qll
qll
EIdx
EIdx
EI
MM ll
8))(
2(
4
1
1 42
00
0 =−−===δ
EI
lll
l
EIdx
EIdx
EI
MM ll
x 3))((
3
1
1 3
00
=−−===δ
q
RA RB
MA
δ0
R
xBR δ
-ql2/2
1
-l

83
Compatibility condition and the final solution
qlREI
lR
EI
qlBB 8
30
38
34
−=→=+ (up)
281
,85
qlMqlR AA −==
Moment Diagram
Deflected shape
4.1.2. The second idea
=
+
00 =θ+θ xAM
-ql2/2
3ql2/8
-ql2/8 -
2
1289
ql
3l/8
+ =
θ0
MA
xAM θ

84
– Rotional Angle at the fixed end
EI
qlqll
EIdx
EIdx
EI
MM ll
241
83
1
1 32
00
0 =×===θ
EI
ll
EIdx
EIdx
EI
MM ll
x 311
3
1
1
00
=××===θ
Compatibility condition and the final solution
23
81
0324
qlMEI
lM
EI
qlAA −=→=+
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

85
4.2 Cantilever Beam with Spring Support
Robin BC (The third type BC)
)()( lkwlwEIV −=′′′−=
Primary structure
wbeam(l)=δspring → spring0 δ−=δ+δ xBR
EIkl
klql
kEI
lEI
ql
Rk
R
EI
lR
EI
qlB
BB 38
31
3
838 3
3
3
4
34
+−=
+−=→−=+
As qlRk B 8
3, →∞→ , and As 0,0 →→ BRk
Deflected Shape for 3
100l
EIk =
δ0
BR
xBR δ
)(lkw

86
4.3 Support Settlement
Primary structure
∆=→∆=δ+δ30
3
l
EIRR BxB
Deflected Shape
∆
RB
xBR δ

87
4.4 Temperature Change
Primary structure
Curvature due to temperature change
dxTTdxTThd )()( 0102 −α−−α=θ
2
212 )(
dx
wd
h
TT
dx
d −=−α
=θ
baxxh
TTw ++
−α−= 212
2
)(
For simple beam, 0)()0( == lww → lh
TTa
2
)( 12 −α=
Comaptibility condition
00 =θ+θ xAM → 032)( 12 =+
−αEI
lMl
h
TTA → EIh
TTM A
)(
2
3 12 −α−=
T1
T2
θ0
MA
xAM θ

88
4.5 Shear Effect
Primary Structures – Shear force diagram
0)()( 000 =δ+δ+δ+δ=δ+δSx
MxB
SMxB RR
GA
fqlql
l
GA
fdx
GA
fdx
GA
VVf
llS
2)1)((
2
2
00
0 ====δ
GA
fll
GA
fdx
GA
fdx
GA
VVf
llSx ====δ )1)(1(
00
2
2
2
2
3
24
7801
0411
8
3
31
41
8
3
3
28
)l/h(.
)l/h(.ql
GAl/fEI
GAl/fEIql
GA/flEI/ql
GA/fqlEI/qlRB +
+−=++−=
++−=
– For retangular section
( ) 222
3
2260
125
30126
12
12
5
6)
l
h(.)
l
h(
.
)v(/Ehbl
/Ebh
GAl
fEI =×
+×=+
=
For 201=
l
h
8
300071
002001
002601
8
3
7801
0411
8
32
2 ql.
.
.ql
)l/h(.
)l/h(.qlRB −=+
+−=++−= (0.07 % error)
1
ql 1
+ +

89
For 101=
l
h
8
300261
007801
010401
8
3
7801
0411
8
32
2 ql.
.
.ql
)l/h(.
)l/h(.qlRB −=+
+−=++−= (0.26 % error)
For 51=
l
h
8
301011
031201
041601
8
3
7801
0411
8
32
2 ql.
.
.ql
)l/h(.
)l/h(.qlRB −=+
+−=++−= (1.0 % 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 RxLxBRL M
EI EI q
ql
Rxθ
Lxθ
1
4
2ql 82ql
q
ql
R0θ L0θ

90
EI
qlqlql
EI
qllqll
EI
dxEI
dxEI
dxEI
MM
ll
lRL
33322
00
2
0
00
48
5)
2416(
1)1
831
4)
2
11(
6(
1
1
1
=+=××+×+=
+=
=θ+θ
EI
ldx
EIdx
EI
dxEI
MM
ll
lRx
Lx
3
2
1
1
00
2
0
=+=
=θ+θ
200
325
qlMRx
Lx
RL
B −=++−=
θθθθ
Deflected shape
1

91
4.7 Fixed-Fixed End Beam
4.7.1. Primary Structure type I
0-th load case (X0)
1st load case (X1)
2nd load case (X2)
ijδ : the deflection in the direction of the unit concentrated load in load case i caused by
load case j. In most of engineering problem, the first subscript denotes results while the
second subscript indicates causes.
EI
qlqll
l
EIdx
EI
MMl
8)
2()(
4
1 42
0
0110 =−×−×==δ
EI
qlqll
EIdx
EI
MMl
6)
2(1
3
1 32
0
0220 −=−××==δ
EI
lll
l
EIdx
EI
MMl
3)()(
3
1 3
0
1111 =−×−×==δ
EI
ll
l
EIdx
EI
MMl
21)(
2
1 2
0
212112 −=×−==δ=δ
EI
ll
EIdx
EI
MMl=××==δ 11
1
0
2222
q
RA RB
MA MB
-ql2/2
M0
1
-l
M1
1 1
M2

92
Compatibility condition (Flexibility equation)
=+−−
=−+→
=δ+δ+δ=δ+δ+δ
026
0238
0
0
21
23
2
2
1
34
22212120
21211110
XEI
lX
EI
l
EI
ql
XEI
lX
EI
l
EI
ql
XX
XX
21ql
X −= , 12
2
2
qlX −=
4.7.2. Primary Structure type II
EI
qlqll
EIdx
EI
MMl
2481
3
1 42
0
0110 =××==δ
EI
qlqll
EIdx
EI
MMl
2481
3
1 32
0
0220 =××==δ
EI
ll
EIdx
EI
MMl
3)1()1(
3
1
0
1111 =−×−×==δ
EI
ll
EIdx
EI
MMl
611
61
δδ
0
212112 =××===
EI
ll
EIdx
EI
MMl
311
3
1
0
2222 =××==δ
M0
M2 1
1
M1
1
1 8
2ql

93
=++
=++→
=δ+δ+δ=δ+δ+δ
03624
06324
0
0
21
3
21
3
22212120
21211110
XEI
lX
EI
l
EI
ql
XEI
lX
EI
l
EI
ql
XX
XX
12
2
21
qlXX −==
Reactions and Moment Diagrams
Deflected Shape
24
2ql 12
2ql−
2
ql 2
ql

94
4.8 3-Span Continuous Beam
4.8.1. Uniform load case
Primary structure
)1(248
13
1
81
3
1
2
1
1
32
2
2
1
2
0
0110 I
I
EI
qlqll
EI
qll
EIdx
EI
MMl+=××+××==δ
)1(248
13
1
81
3
1
2
1
1
32
1
2
2
2
0
0220 I
I
EI
qlqll
EI
qll
EIdx
EI
MMl+=××+××==δ
)1(3
113
111
3
1
2
1
121
2
0
1111 I
I
EI
ll
EI
l
EIdx
EI
MMl+=××+××==δ
220
212112 6
116
1δδ
EI
ll
EIdx
EI
MMl =××===
)1(3
113
111
3
1
2
1
112
2
0
2222 I
I
EI
ll
EI
l
EIdx
EI
MMl+=××+××==δ
EI1 EI2
q
EI1
M0
1
M1
1
M2

95
=++++
=++++→
=δ+δ+δ=δ+δ+δ
0)1(36
)1(24
06
)1(3
)1(24
0
0
22
1
11
22
1
1
3
22
12
1
12
1
1
3
22212120
21211110
XI
I
EI
lX
EI
l
I
I
EI
ql
XEI
lX
I
I
EI
l
I
I
EI
ql
XX
XX
2
1
2
1
221
5.11
1
8
1
I
II
I
qlXX+
+−==
In case 21 II = , 2
21 101
qlXX −==
4.8.2. Complicated Load Case
Primary structure
M0
EI EI
q
EI

96
EI
qlqll
EIdx
EI
MMl
2481
3
1 32
0
0110 =××==δ
EI
ql
EI
ql
EI
qlqll
EI
qll
EIdx
EI
MMl
48
5
162441)
2
11(
6
1
81
3
1 333222
0
0220 =+=××++××==δ
=++
=++→
=δ+δ+δ=δ+δ+δ
03
2
648
5
063
2
240
0
21
3
21
3
22212120
21211110
XEI
lX
EI
l
EI
ql
XEI
lX
EI
l
EI
ql
XX
XX
2
1 401
qlX −= , 22 406
qlX −=
Compatibility Condition (Flexibility Equation) in General
=
∆=δ+δn
jijiji X
10

97
Chapter 5
Analysis of Statically Indeterminate Trusses
1 1
1
21− 2
1−
21−

98
5.1 Various Types of Trusses
Determinate Truss
Externally Indeterminate Truss
Internally Indeterminate Truss
Mixed Indeterminate Truss

99
5.2 A Simple Truss
5.2.1 Method - I
+
P
X
P
P
=

100
Primary structure
At L1
0
75.06.0
25.1
18.0
1
31
3
==+
==
F
FF
F
F
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
L1 L2
U1 U2

101
At L2
75.0 , 25.1
18.0
06.0
45
5
54
−===
=+
FF
F
FF
At U1
1 , 75.0
08.0
06.0
21
52
51
−=−==+=+
FF
FF
FF
1
-1.0
1.25
1.25
-0.75
1
-0.75
F4 F5
1
F1
F2
F5

102
Axial force table for primary structure
Mem 0F xF A L LEA
FF x00 =δ
LEA
Fxx
2
=δ
1 0 -0.75 0.5 0.75L 0 LEA5.0
75.0 3
2 -P -1.0 0.5 L EA
PL
5.0 L
EA5.01
3 1.25 P 1.25 1.0 1.25L EA
PL325.1
EA
L325.1
4 -0.75 P -0.75 0.5 0.75L EA
PL
5.075.0 3
LEA5.0
75.0 3
5 0 1.25 1.0 1.25L 0 EA
L325.1
EA
PL79.4
EA
L59.7
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

103
5.2.2 Method - II
P
+
P
X
X
=

104
EA
XLX 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
EA
XL

105
Mem 0F xF A L LEA
FF x00 =δ LEA
FF xxx =δ
1 0 -0.6 0.5 0.75L 0 EA
L54.0
2 -P -0.8 0.5 L EA
PL6.1 EA
L28.1
3 1.25 P 1.0 1.0 1.25 L EA
PL56.1 EA
PL25.1
`4 -0.75 P -0.6 0.5 0.75 L EA
PL68.0
EA
PL54.0
5 - - 1.0 - - -
EA
PL84.3 EA
L61.3
EA
XLX x −=δ+δ0 → XEA
LX
EA
L
EA
PL 25.161.384.3 −=+
PPX 79.086.4
84.3 −==
PXH 63.08.02 −==

106
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 - 2P - 2P - 22 P
1
1
1
2
1− 2
1−
2
1−
2
1−

107
Mem F 0 xF EAL
LEA
FF x00 =δ LEA
FF xxx =δ
1 P 0 1 0 0
2 P2− 0 2 0 0
3 P 2
1− 1 2P−
2
1
4 P2 2
1− 1 22P−
2
1
5 P2− 1 2 P2− 2
6 - - 2 - -
7 P− 21− 1
2
P
2
1
8 P3 2
1− 1 23P−
2
1
9 P2 0 1 0 0
10 P22− 0 2 0 0
-5.54P 3.41
LEA
XXx 20 −=δ+δ
PXXXP 15.141.141.354.5 =→−=+−
X

108
Temperature change and fabrication error
)22(0 fx LTLEA
XX ∆+∆α+−=δ+δ
In case of no external loads
)2(21.0)2(82.4 ff LTL
EAXLTX
EA
L ∆+∆−=→∆+∆−= αα
5.4 Truss with Two Hinge Supports
Primary structure and compatibility condition
0
2
22212120
121211110
=δ+δ+δ
−=δ+δ+δ
XX
LEA
XXX
3P 2P P
X1
1
2
3
4
5
6
7
8
9

structural analysis istrana.snu.ac.kr/lecture/struct1_2019/notes/19_note(ch1... · 2019-05-03 ·...

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Buckling of Rigid Frames – I - UMass of Rigid Frames – I ... Slope Deflection Method • Slope deflection method is a displacement-based analysis for indeterminate structures

Chapter 3 Trusses - موقع المهندس حماده شعبان 3 final2 (2).pdfb + r 2 j statically indeterminate degree of indeterminacy = b + r – 2 j ،ٹپي ٽڃ٫پ ٽ ٲأ

Semântica de frames e análise crítica do discurso: da ...lise-do- · Semântica de frames e análise

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Frames Com frames você pode inserir mais de um documento dentro da mesma tela do navegador. Você pode fazer frames horizontais ou verticais, que determinarão

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Indeterminate Propositions in Indeterminate Propositions in Prior Analytics I.41 167 deals with exactly

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FACULTY OF TECHNOL OGY - Engineering Classes | ... and Twisting Moment in various statically indeterminate structures such as beams, ... Fixed Beams, Application of the ... direct

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Structural Analysis Structural Analysis - Classical and Matrix Approach by Jack C. McCormac and R. E. Elling - Elementary Structural Analysis 4th Ed. by C. H. Norris - Statically Indeterminate

Analysis of Statically Indeterminate .– Indeterminate to the first degree ... redundant & solve