theory of structures 07
DESCRIPTION
Theory of Structures 07TRANSCRIPT
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7. Deflection: Work-Energy Method 1
7. Deflections of Truss, Beams, and Frames
: Work-Energy Method
Geometric Method– Direct Integration Method– Moment – Area Method– Conjugate – Beam Method (Elastic Weight Method)
Work – Energy MethodAdvantage : Applied to beams, trusses, and frame Disadvantage : Computing only at one point
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7. Deflection: Work-Energy Method 2
7.1 Work
• Work (Done) = Force ⅹ Displacement
∆
P
Disp.
Force
∆
P
P∆= PddW
∫∆
∆=0
PdW
d ∆
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7. Deflection: Work-Energy Method 3
• Linear Elastic Deformation
Disp.
ForceP
∆
∆= PW21
• Constant Force during a deformation
Disp.
Force
∆
P ∆= PW
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7. Deflection: Work-Energy Method 4
• Work (Done) = Moment ⅹ Rotation
θMddW =
∫=θ
θ0
MdW
θMW21
= --- Linear Elastic Defoemation
θMW = --- Constant Moment Magnitude
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7. Deflection: Work-Energy Method 5
Homework 1 (5 Points)
• Summarize the unit system of force, displacement, work and energy
- USCS (U. S. Customary System)
- SI (International System)
- MKS (Meter-Kilogram-Second)
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7. Deflection: Work-Energy Method 6
7.2 Principle of Virtual Work
• External Work : Work done by External ForceInternal Work : Work done by Internal Force
• External Forces : Loads, ReactionsInternal Forces : Axial Forces, Shear Forces, Bending Moments
P
Ax
Ay Cy
C
BA
Axial Force(축력)
Shear Force(전단력)
Bending Moment (휨모멘트)• External Work = Internal Work
We = Wi
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7. Deflection: Work-Energy Method 7
• Virtual Work
Virtual Work = Work done by Virtual Force= Virtual Force × Real Displacement
Virtual Work = Work done due to Virtual Displacement= Real Force × Virtual Displacement
Virtual Work = Virtual Force × Virtual Displacement
• Virtual External Work = Work done by Virtual External Force
Virtual Internal Work = Work done by Virtual Internal Force
• Principle of Virtual Work (John Bernoulli, 1717)
Virtual External Work = Virtual Internal Work
Wve = Wvi
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7. Deflection: Work-Energy Method 8
B
A
Cθ2
θ1
P
∆
θ2
θ1
B
A
C Pv Pv
FvBC
FvAC
Cθ2
θ1
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7. Deflection: Work-Energy Method 9
Virtual External Force : Pv
Virtual Internal Force : FvAC , FvBC
Pv
FvBC
FvAC
Cθ2
θ1
0=Σ xF0coscos 21 =−− θθ vBCvACv FFP
21 coscos θθ vBCvACv FFP +=
0sinsin 21 =+− θθ VBCVAC FF0=Σ yF
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7. Deflection: Work-Energy Method 10
B
A
Cθ2
θ1
P
∆
θ2
θ1
B
A
C Pv
FvBC
FvAC
Virtual External Work = Virtual External Force × Real External Displacement
Wve = Pv ∆
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7. Deflection: Work-Energy Method 11
B
A
Cθ2
θ1
P
∆
θ2
θ1
B
A
C Pv
FvBC
FvAC
Virtual Internal Work = Virtual Internal Force × Real Internal Displacement
∆
θ2
θ1
δBC
δAC
θ2
θ1
δAC = ∆ cosθ1
δBC = ∆ cosθ2
Wvi = FvACδAC + FvBCδBC
= FvAC ∆ cosθ1 + FvBC∆ cosθ2
= (FvAC cosθ1 + FvBCcosθ2) ∆
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7. Deflection: Work-Energy Method 12
Wve = Pv ∆
Wvi = (FvAC cosθ1 + FvBCcosθ2) ∆21 coscos θθ vBCvACv FFP +=
Virtual External Work = Virtual Internal Work
Wve = Wvi
Σ (Virtual External Force × Real External Force)= Σ (Virtual Internal Force × Real Internal Force)
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7. Deflection: Work-Energy Method 13
7.3 Deflections of Trusses by the Virtual Work Method
Real System
2P1P
j
C
D
AB
Determination of the vertical deflection at joint B
FjmemberofForceAxial =
)( MaterialsofMechnicsFromFAEL
=δ
δ=jmemberofnDeformatioAxial
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7. Deflection: Work-Energy Method 14
Virtual System
- Virtual External Force Pv = 1
- 처짐을 구하고자 하는 절점에, 처짐의 방향으로
1
∆
2P1P
1
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7. Deflection: Work-Energy Method 15
1∆
2P1P
∑ ×= )Re(Wve ntDisplacemeExternalalForceExternalVirtual
ƥ=1
)Re(Wvi ntDisplacemeInternalalForceInternalVirtual ×=∑δ∑= vF
∑= FAELFv∑=∆ δvF
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7. Deflection: Work-Energy Method 16
∑=∆ FAELFv
1ForceExternalVirtualtodueForceInternalVirtualFv =
ForceExternalaltodueForceInternalalF ReRe=
• Deflections due to Temperature Changes
ChangeeTemperaturtoduentDisplacemeInternalalRe=δ
LT )(∆=α
ExpansionThermaloftCoefficien=α
∑=∆ δvF ∑ ∆= LTFv )(α
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7. Deflection: Work-Energy Method 17
Example 7.1
Determine the horizontal deflection at joint G.
k40
k20 G
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7. Deflection: Work-Energy Method 18
(1) Analysis of Real System
k40
k20G
FForceIntervalalRe
(2) Analysis of Virtual System
G1
vFForceIntervalVirtual
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7. Deflection: Work-Energy Method 19
(3) Determination of Deflection
∑= FALF
E v1∑=∆ F
AELFv
FFAL v)/(member )(inL )( 2inA )(kFv )(kF
∑
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7. Deflection: Work-Energy Method 20
Example 7.2
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7. Deflection: Work-Energy Method 21
Example 7.3
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7. Deflection: Work-Energy Method 22
Example 7.4
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7. Deflection: Work-Energy Method 23
7.4 Deflections of Beams by the Virtual Work Method
Real System
∆
P
L
ω (x)
B
Virtual System for determining ∆
B
1
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7. Deflection: Work-Energy Method 24
∑ ×= )Re(Wve ntDisplacemeExternalalForceExternalVirtual
ƥ=1
)Re(Wvi ntDisplacemeInternalalForceInternalVirtual ×=∑
xdx ∆
L
ω (x)
B
P
θ θθ d+
vM vM
dx
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7. Deflection: Work-Energy Method 25
θ θθ d+
vM vM
dx
θθθ vvvi MdMdW −+= )( θdM v=
)( DeflectionBeamofEquationalDifferentiFromdxEIMd =θ
dxEIMMdW vvi =
∫=L
vvi dxEIMMW
0
∫=∆L
v dxEIMM
0
1ForceExternalVirtualtodueMomentBendingVirtualM v =
ForceExternalaltodueMomentBendingalM ReRe=
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7. Deflection: Work-Energy Method 26
Virtual System for determining θ
1
θ•=1veW
∫=L
vvi dxEIMMW
0
∫=L
v dxEIMM
0θ
1MomentExternalVirtualtodueMomentBendingVirtualM v =
ForceExternalaltodueMomentBendingalM ReRe=
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7. Deflection: Work-Energy Method 27
θ•=1veW
∫=L
vvi dxEIMMW
0
∫=L
v dxEIMM
0θ
Virtual Internal Work = Virtual Internal Work done by Bending Moment
+ Virtual Internal Work done by Shear Force
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7. Deflection: Work-Energy Method 28
Homework 2 (10 Points)
• Derive the Force-Deformation Relatioship of Truss
FAEL
=δ
• Derive the Force-Deformation Relatioship of Beam
MEIdx
d 1=
θ
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7. Deflection: Work-Energy Method 29
Example 7.5
AB
EI= CONSTANT
L
w
Determine the slope and deflection at point A.
Real Systemw
x
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7. Deflection: Work-Energy Method 30
w
x
)(xw
wLxxw =)(
=)(xM −L
wx6
3
−=)31( x)
21)((x
Lxw
Virtual System for Determining Slope at A
x
1
1)( =xMv
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7. Deflection: Work-Energy Method 31
∫=L v
A dxEI
MM0
θ ∫ −=−=L
EIwLdX
Lwx
EI0
33
24)
6)(1(1
(-) : Virtual Force의 방향과 반대
EIwL
A 24
3
=θ
Virtual System for Determining Deflection at A
x
1
xMv −=
∫ =−
−=L
EIwLdX
Lwxx
EI0
43
30)
6)((1
∫=∆L v
A dXEI
MM0
(+) : Virtual Force의 방향과 동일
EIwL
A 30
4
=∆
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7. Deflection: Work-Energy Method 32
Example 7.6
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7. Deflection: Work-Energy Method 33
Example 7.7
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7. Deflection: Work-Energy Method 34
7.5 Deflections of Frames by the Virtual Work Method
• Internal forces of trusses → Axial Forces
• Internal forces of beams → Bending Moments
→ Shear Forces
• Internal forces of frames → Bending Moment
→ Shear Forces
→ Axial Forces
∑= AEFLFW vvi ∫∑+ dx
EIMM v
∫∑∑ +=∆ dxEIMM
AEFLF vv ∫∑=∆ dx
EIMM v
∫∑= dxEIMM vθ∫∑∑ += dx
EIMM
AEFLF vvθ
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7. Deflection: Work-Energy Method 35
Example 7.8
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7. Deflection: Work-Energy Method 36
Example 7.9
15 ft
2 k/ft
10 k
10 ft 10 ft
A
B C
D
E E = 29,000 ksi
I = 1,000 in4
A = 35 in2
Determine the lateral deflection at point C.
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7. Deflection: Work-Energy Method 37
1. Real System 2 k/ft
10 k
Ay
AxBx
By
)(67.1 →= kAx
)(50.12 ↑= kAy
)(67.11 ←= kBx
)(50.27 ↑= kBy
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7. Deflection: Work-Energy Method 38
2 k/ft
10 k
12.50 k
1.67 k 11.67 k
27.50 k
A
B C
D
E
(ft) Range
scoordinate XSegment (k)F ft)(kM −
Origin
x67.1−50.12−150 −AAB67.11− 25.1205.25 xx −+−200 −BBC
x67.11−200 − 50.27−CD D
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7. Deflection: Work-Energy Method 39
2. Virtual System1 k
Ay
AxBx
By
)(50.0 ←= kAx
)(75.0 ↓= kAy
)(50.0 ←= kBx
)(75.0 ↑= kBy
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7. Deflection: Work-Energy Method 40
1 k
0.75 k
0.50 k 0.50 k
0.75 k
A
B C
D
E
(ft) Range
scoordinate XSegment (k)Fv ft)(kM v −
Origin
x50.075.0150 −AABx75.050.7 −50.0200 −BBC
x50.0−200 − 75.0−CD D
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7. Deflection: Work-Energy Method 41
3. Calculation of Deflection
∑= FAELFv
[ ])50.27(15)75.0()67.11(205.0)50.12(1575.01−××−+−××+−××=
AE
AEftk )(05.52 2 −
= 000,29351205.52
××
= )(00062.0 →= in
1C∆
∑∫= dxEIMM v
−−+−+−−+−= ∫∫∫
15
0
20
0
215
0)67.11)(5.0()5.1205.25)(75.05.7()67.1(5.01 dxxxdxxxxdxxx
EI
2C∆
EIftk )(375,9 32 −
=1000000,2912375,9 3
××
= )(55862.0 →= in
21 CCC ∆+∆=∆ )(55924.0 →= in
C
C
∆∆ 1
55924.000062.0
= 00110.0=
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7. Deflection: Work-Energy Method 42
7.6 Conservation of Energy and Stain Energy
• Work
• Energy
- Capacity for doing work
Principle of Conservation of Energy
Work done by external forces = Work done by internal forces
External Work = Internal Work
ie WW = Strain Energy, UUWe =
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7. Deflection: Work-Energy Method 43
Strain Energy of Trusses
External Forces → Internal Forces (Axial Forces)
F
FAEL
=δ
AELFFU j 22
1 2
== δ
AELFU
2
2
Σ=
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7. Deflection: Work-Energy Method 44
Strain Energy of Beams
External Forces → Internal Forces (Bending Monents)
M
dxEIMd =θ
dxEI
MMddU22
1 2
== θ
dxEI
MUL
∫= 0
2
2
Strain Energy of Frames
dxEI
MAE
LFU ∫∑ ∑+=22
22
dxEI
MU ∫∑= 2
2
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7. Deflection: Work-Energy Method 45
7.7 Castgliano’s Second Theorem
Alberto Castigliano, 1873
iiP
U∆=
∂∂
iiM
Uθ=
∂∂
= Pi가 작용하는 점에서 Pi의 작용방향으로 발생하는 처짐i∆
iθ = Mi가 작용하는 점에서 Mi의 작용방향으로 발생하는 처짐각
Application to Trusses
)2
(2
AELF
PΣ
∂∂
=∆
AEFF
PF ∂
=∂∂ 2
2
AEFL
PF∂∂
Σ=∆
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7. Deflection: Work-Energy Method 46
Application to Beams
dxEI
MP
L
∫∂∂
=∆0
2
2dX
EIM
PML
∫ ∂∂
=0 2
dxEIM
MML
∫ ∂∂
=0 2
θ
Application to Frames
dXEIM
PM
AEFL
PF
∫∑ ∑ ∂∂
+∂∂
=∆2
dXEIM
PM
∫∑ ∂∂
=∆2
dXEIM
MM
∫∑ ∂∂
=2
θdXEIM
MM
AEFL
MF
∫∑ ∑ ∂∂
+∂∂
=2
θ
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7. Deflection: Work-Energy Method 47
Example 7.10
2k/ft 12k
30’ 10’
CBA
EI = constant
E = 29,000 ksi
I = 2,000 in4
Evaluate the deflection at point C.
= P
330 P−
3430 P
+
2k/ft 12k
CBA
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7. Deflection: Work-Energy Method 48
= P
330 P−
3430 P
+
2k/ft 12k
CBA
x x
For segment AB (0ft – 30ft)
xP
M )330( −=
3x
PM
−=∂∂
2
330 xPxx −−=xx
212 ⋅−
For segment BC (0ft – 10ft)
PxM −=
xPM
−=∂∂
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7. Deflection: Work-Energy Method 49
∫ ∂∂
=∆L
C dxEIM
PM
0
])12)(()3
1230)(3
([1 10
0
30
0
2 ∫∫ −−+−−−= dxxxdxxxxxEI
kftk −
ftk −• ft•EI
6500−= 3ftk −=
EIftk
C
36500 −−=∆
200029000126500 3
××
−= in194.0−=
)(194.0 ↑=∆ inC
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7. Deflection: Work-Energy Method 50
Example 7.11
1.5k/ft
40kB
A
CD
30ft
12ft
12ft
Evaluate the rotation at point C.
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7. Deflection: Work-Energy Method 51
1.5k/ft
40kB
A
CD
M
305.6 M−
305.38 M+
40
MM∂∂
,M
∫∑ ∂∂
=∆ dxEIM
MM
C 0=← M
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7. Deflection: Work-Energy Method 52
Example 7.12
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7. Deflection: Work-Energy Method 53
Homework 3 (40 Points)
Solve the following problems.7.1 ~ 7.3 : 1 problem7.4 ~ 7.7 : 1 problem7.8 ~ 7.10 : 1 problem7.11 ~ 7.13 : 1 problem7.14 ~ 7.15 : 1 problem7.18 ~ 7.23 : 2 problems7.24 ~ 7.26 : 1 problem7.27 : 1 problem7.28 ~ 7.40 : 2 problems7.41 ~ 7.45 : 1 problem7.46 ~ 7.50 : 1 problem7.50 ~ 7.54 : 1 problem
14 problems
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7. Deflection: Work-Energy Method 54
Quiz 1 (60 Points)
1. 다음 용어에 대해서 간단히 답하시오. (20 Points)
(1) 평형방정식 (Equations of Equilibrium)
(2) 외력 & 내력 (External Force & Internal Force)
(3) 자유물체도 (Free-Body Diagram)
(4) 정정 구조물 & 부정정 구조물 (Determinate Structures & Indeterminate Structures)
(5) Roller, Hinged Support, & Fixed Support
2. 다음 구조물의 반력을 구하고, 전단력도와 휨모멘트도를 그리시오. (20 Points)
AB C ED
50 kN 10 kN/m
3 m 3 m 4 m 4 m
200 kN-m
2 m
F
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7. Deflection: Work-Energy Method 55
3. 다음 트러스 구조물의 부재력을 구하시오. (20 Points)
9 ft
4 at 12ft = 48ft
A
B C D
E
F G H
20k 20k 20k
3
4
5