Theory of Structures 07

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


7.1 Work

• Work (Done) = Force ⅹ Displacement

∆

P

Disp.

Force

∆

P

P∆= PddW

∫∆

∆=0

PdW

d ∆


• Linear Elastic Deformation

Disp.

ForceP

∆

∆= PW21

• Constant Force during a deformation

Disp.

Force

∆

P ∆= PW


• Work (Done) = Moment ⅹ Rotation

θMddW =

∫=θ

θ0

MdW

θMW21

= --- Linear Elastic Defoemation

θMW = --- Constant Moment Magnitude


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)


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


• 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


B

A

Cθ2

θ1

P

∆

θ2

θ1

B

A

C Pv Pv

FvBC

FvAC

Cθ2

θ1


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


B

A

Cθ2

θ1

P

∆

θ2

θ1

B

A

C Pv

FvBC

FvAC

Virtual External Work = Virtual External Force × Real External Displacement

Wve = Pv ∆


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) ∆


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)


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


Virtual System

- Virtual External Force Pv = 1

- 처짐을 구하고자 하는 절점에, 처짐의 방향으로

1

∆

2P1P

1


1∆

2P1P

∑ ×= )Re(Wve ntDisplacemeExternalalForceExternalVirtual

∆•=1

)Re(Wvi ntDisplacemeInternalalForceInternalVirtual ×=∑δ∑= vF

∑= FAELFv∑=∆ δvF


∑=∆ FAELFv

1ForceExternalVirtualtodueForceInternalVirtualFv =

ForceExternalaltodueForceInternalalF ReRe=

• Deflections due to Temperature Changes

ChangeeTemperaturtoduentDisplacemeInternalalRe=δ

LT )(∆=α

ExpansionThermaloftCoefficien=α

∑=∆ δvF ∑ ∆= LTFv )(α


Example 7.1

Determine the horizontal deflection at joint G.

k40

k20 G


(1) Analysis of Real System

k40

k20G

FForceIntervalalRe

(2) Analysis of Virtual System

G1

vFForceIntervalVirtual


(3) Determination of Deflection

∑= FALF

E v1∑=∆ F

AELFv

FFAL v)/(member )(inL )( 2inA )(kFv )(kF

∑


Example 7.2


Example 7.3


Example 7.4


7.4 Deflections of Beams by the Virtual Work Method

Real System

∆

P

L

ω (x)

B

Virtual System for determining ∆

B

1


∑ ×= )Re(Wve ntDisplacemeExternalalForceExternalVirtual

∆•=1

)Re(Wvi ntDisplacemeInternalalForceInternalVirtual ×=∑

xdx ∆

L

ω (x)

B

P

θ θθ d+

vM vM

dx


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=


Virtual System for determining θ

1

θ•=1veW

∫=L

vvi dxEIMMW

0

∫=L

v dxEIMM

0θ

1MomentExternalVirtualtodueMomentBendingVirtualM v =

ForceExternalaltodueMomentBendingalM ReRe=


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



• Derive the Force-Deformation Relatioship of Truss

FAEL

=δ

• Derive the Force-Deformation Relatioship of Beam

MEIdx

d 1=

θ


Example 7.5

AB

EI= CONSTANT

L

w

Determine the slope and deflection at point A.

Real Systemw

x


w

x

)(xw

wLxxw =)(

=)(xM −L

wx6

3

−=)31( x)

21)((x

Lxw

Virtual System for Determining Slope at A

x

1

1)( =xMv


∫=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

=∆


Example 7.6


Example 7.7


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


Example 7.8


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.


1. Real System 2 k/ft

10 k

Ay

AxBx

By

)(67.1 →= kAx

)(50.12 ↑= kAy

)(67.11 ←= kBx

)(50.27 ↑= kBy


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


2. Virtual System1 k

Ay

AxBx

By

)(50.0 ←= kAx

)(75.0 ↓= kAy

)(50.0 ←= kBx

)(75.0 ↑= kBy


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


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=


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 =


Strain Energy of Trusses

External Forces → Internal Forces (Axial Forces)

F

FAEL

=δ

AELFFU j 22

1 2

== δ

AELFU

2

2

Σ=


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


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=∆


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


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


= 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

−=∂∂


∫ ∂∂

=∆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


Example 7.11

1.5k/ft

40kB

A

CD

30ft

12ft

12ft

Evaluate the rotation at point C.


1.5k/ft

40kB

A

CD

M

305.6 M−

305.38 M+

40

MM∂∂

,M

∫∑ ∂∂

=∆ dxEIM

MM

C 0=← M


Example 7.12



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


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


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

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Theory of Structures 07