Chapter 6f

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VIRTUAL WORK

& RECIPROCAL PRINCIPLESCHAPTER 6



OUTLINE

Introduction

External and internal work

Energy method

Principle of virtual work method

Virtual work method for truss, beam,frame



INTRODUCTION

1. Geometric method

Double integration method

Moment- area method

Conjugate beam method

2. Energy method Virtual work

Castigliano’s theorem

Simple loading

Beam

Complicated loading

Truss

Frame



INTRODUCTION

Most energy method: based on conservation of energy principle

Work done by external forces acting on structure (Ue) is transformed

into internal work or strain energy (Ui)

Ue

= Ui



EXTERNAL WORK- FORCE

Work = Force x Displacement

Gradually increasing load

=

= 12∆



EXTERNAL WORK- FORCE

Due to another force;

=

= ∆′



EXTERNAL WORK- MOMENT

=

= 12

= ′

Moment:

Moment is

gradually applied:

Moment is already applied to thestructure, and other loadings further

distort the structure by amount θ’:



STRAIN ENERGY- AXIAL FORCE

Hooke’s Law:

Stress:

Strain:

= = /

= ∆/∆=

= 12∆=

2



STRAIN ENERGY- BENDING

From beam theory;

=

= 12 =

2

=

2



PRINCIPLE OF WORK & ENERGY

P

L

x

VM

P

=

12∆= 2

1

2∆=

2∆=

3

=



PRINCIPLE OF VIRTUAL WORK

Provide generalization of equilibrium and compatibility concepts

Apply to linear and non-linear structures and solids under small and

large displacements

Unit load method (or dummy load method) is derived from principleof virtual work – powerful tool to evaluate displacements




External and internal load: related through equations of equilibrium

External and internal displacement: related through equations of

compatibility of the displacements

∆

Work ofExternal Loads

Work ofInternal Loads

=




= 12∆ + 1 ∆ = +

Real Work

1 ∆ = Virtual Loadings

Real Displacements

Similarly;

1 = Virtual Loadings

Real Displacements



VIRTUAL WORK FOR TRUSS

- EXTERNAL LOAD

Virtual Load

Real Load


Real Displacements

=

=

1 ∆ =

1 = external virtual load acting in the same jointin the stated direction of ∆

∆ = external joint displacement caused by real

load

n = internal virtual normal force caused by

external virtual load

N = internal normal force caused by real load




- TEMPERATURE

1 ∆ =

Virtual Loadings

Real Displacements

= ∆ =

1 ∆ = ∆ ∆ = external joint displacement caused bytemperature change

α = coefficient of thermal expansion of member

∆T = change in temperature of member




- FABRICATION ERROR

1 ∆ =

Virtual Loadings

Real Displacements

= ∆ =

1 ∆ = ∆ ∆ = external joint displacement caused byfabrication error

∆L = difference in length of the member from its

intended size as caused by fabrication error




- COMBINED EFFECTS

1 ∆ = +∆ + ∆



EXAMPLE 1

The cross sectional area of each truss member is E = 200 GPa,A = 400 mm2.

(a) vertical displacement at joint C due to 4 kN force(b) if no load act on truss, find

the vertical displacement atjoint C if member AB is tooshort by 5 mm

(c) If both (a) and (b) areaccounted for, find thevertical displacement atjoint C

A B

C4 kN

4 m 4 m

3 m



A B

C

1 kN

3 m

0.5 kN 0.5 kN

0.667

Virtual Force n: since vertical displacement at joint C is tobe determined, a 1kN vertical load at joint C is to be

placed. All other external forces MUST BE REMOVED.Find the internal forces using method of joint.



AB

C4 kN

3 m

1.5 kN 1.5 kN

2

4 kN

Real force N : The internal force in each member iscalculated using method of joint.



Calculate the displacement due to 4 kN force

Member ni (kN) Ni (kN) Li (m) ni (NiLi) (kN2.m)

AB

AC

CB

0.667

-0.833

-0.833

2

2.5

-2.5

8

5

5

10.67

-10.41

10.41

Σ ni (NiLi) 10.67

EXAMPLE 1

1 ∆ =

1 ∆ = 1 = 1

80000 ×10.67

∆= 0.133 *As the answer is +ve, the displacement is in

the direction of virtual load (ie; downward)



EXAMPLE 1

(b) Vertical displacement at joint C if member AB is too short by 5 mm

A B

C

5 mm



EXAMPLE 1

(b) Vertical displacement at joint C if memberAB is too short by 5 mm

Virtual load n has been calculated Vertical displacement at joint C due to fabrication error;

1 ∆ = ∆

Member AB

Difference in length ofthe member, -5 mm (short)

∆= 0.667 0.005 = 0.0033

∆= 3.33

*Ensure consistent unit

*As the answer is -ve, the displacement is in the

opposite direction of virtual load

A B

C

1 kN

0.5 kN 0.5 kN

0.667



EXAMPLE 1

(c) Find the vertical displacement at joint C if both (a) & (b) are tooccur

1 ∆ = +∆∆= 0.133 3.33

∆= 3.2



EXAMPLE 2

Calculate the vertical deflection

at joint C due to load and

temperature change. Take:

E = 200 GPa

α = 0.6 x 10-5/oF

ΔTAD = 120oF

AAD = 1300 mm2

AAB = 1300 mm2

ABC = 1300 mm2

ACD = 1300 mm2

AAC = 900 mm2

60 kN

80 kN

8 m

AB

CD

6 m



EXAMPLE 3

Determine the vertical displacement at joint C under the appliedloading. Take A = 350 mm2 and E = 200 GPa

4 kN

A

4 kN

B CD

EF

10 m

10 m 10 m 10 m



VIRTUAL WORK FOR BEAMS & FRAMES

- VERTICAL DISPLACEMENT

1 ∆ =

Virtual Loadings

Real Displacements

= =




- VERTICAL DISPLACEMENT


Real Displacements1 ∆ =

()

Real Load Virtual Unit Load

u=m




- SLOPE

Real Load Virtual Unit Couple




- SLOPE

1 =

Virtual Loadings

Real Displacements

1 =

()

= =

=



EXAMPLE 4

Find the slope and displacement at point B.E=200 GPa, I=250(106) mm4

3 kN/m

5 m

A B



Establish appropriate coordinate x

Vertical displacement at B → apply virtual unit load at B

1 kN

x

A B

1 kN

x

m

= 0;

= + 1 = 0



Real load

Use the same coordinate of x

3 kN/m

x

m

= 0;

+ 3 2 = 0

3 kN/m

=

3

2



Solve to find ∆B

1 ∆ =

()

=

3/2 =

3/2

= 234.4

1 ∆ = 234.4

200 10 / × 250 10−

∆= 4.69

*Ensure units are consistentto avoid confusion



1 =

()

= 0.00125

EXAMPLE 4

Slope at point B (try on your own)



EXAMPLE 5

Determine the slope and displacement of point B.E=200 GPa, I=60(106) mm4

5 kN

A

B

C D

14 kNm

2 m 2 m 3 m



Establish appropriate coordinate x

Vertical displacement at B → apply virtual unit load at B

1

A

B

C D

x1

x2

x3

1 = 0.5 = 0.5

1 ∆ =

()



Real load

Use the same coordinate of x

5 kN

A

B

C D

x1

x2

x3

14 kNm

14

= 14

= 6

6 kN1 kN

12



Solve to find ∆B

1 ∆ =

()

=

0.5 14 +

0.5 6 +

0

= 20.667

1 ∆ = 20.667

200 10 / × 60 10−

∆= 1.72

*Ensure units are consistentto avoid confusion



Slope at point B (try on your own)

1 =

()

= 0.000194

EXAMPLE 5



EXAMPLE 6

(a) Determine the slope and horizontal displacement of point C

(b) Draw the bending moment diagram and deflected curve

E=200 GPa, I=200(106) mm4

5 m

A

B

4 kN

6 m

2 kN/m

C

1.5 EI

EI



QUESTIONS?

Chapter 6f

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