PDF Conjugate structure I

Conjugate Structure Analogy

Introduction

¾ Perform deformation analysis of flexure-dominated structures

ƒ Beams

ƒ Frames

¾ Provide equations to determine

ƒ Displacement

ƒ Rotation

Basic Assumptions

¾ Small displacement (u,v) and small rotation (θ)

ƒ |u/L|, |v/L|, |θ| << 1; L ~ characteristic dimension of the structure

ƒ Rotation is approximated by θ = dv/dx

ƒ Curvature is approximated by κ = d²v/dx²

¾ Kinematics of the cross section

ƒ Plane section remains plane

ƒ No shear deformation Æ Plane section always normal to Neutral Axis

NA NA

Deformed state Undeformed state

Y

Deformed state X

Undeformed state

Deformed state

X Y

Undeformed state

Basic Assumptions (Cont.)

¾ No axial deformation (axially rigid members) Æ no change in length of all members

ƒ L = Lo

ƒ uA = uB

¾ Material behavior

ƒ Linearly elastic, i.e. linear stress-strain relation

ƒ Isotropic, i.e. material properties are directional independent

¾ Equilibrium of the structure

ƒ Equilibrium equations are set up on undeformed state

ε σ

E = Young Modulus 1

A

B

L Lo

uA

vA

uB

vB Deformed state

Undeformed state

Conjugate Structure Analogy for Straight Segment

¾

Moment Area Equations in Local Coordinate System

ƒ Moment Area Equations

ƒ Length Constraint Equation

¾

Coordinate Transformation

^{AB AB}

A AB

B −θ = A θ

( )

B

AB AB AB A AB A AB

B v L x

v − +θ = A

u uA^AB 0

AB

B − =

LAB

A

B

x y

x B

x A

AB

uB

M/EI diagram

Centroid

AB B AB A θ

θ , = Rotations at A and B

AB B AB A v

v , = Transverse components of displacement at A and B

AB B AB

A u

u , = Longitudinal components of displacement at A and B

AB

vB AB

θB

AB

uA AB

vA AB

θA

P

x y

P´

X Y

P P=θ Θ

u P

v P

UP

V P

{x,y} ~ Local coordinate system {uP, vP, θP}

{UP, VP, ΘP}

~ Displacement and rotation in local coordinate system {X,Y} ~ Global coordinate system

~ Displacement and rotation in global coordinate system

U_P= u_Pcosφ−v_Psinφ φ

φ + φ

= _Psin _Pcos

P u v

V

P P= θ Θ I

II

III

AAB = Area of M/EI diagram of the segment AB

¾

Moment Area Equations in Global Coordinate System

ƒ Coordinate transformation at end A

ƒ Coordinate transformation at end B

ƒ Substituting equations (VI) and (IX) into equation (I) leads to

ƒ By first combining the equations (II) and (III) such that (III)cosφ – (II)sinφ, and then using equations (IV), (VI) and (VII), we obtain

ƒ By first combining the equations (II) and (III) such that (III)sinφ + (II)cosφ, and then using equations (V), (VI) and (VIII), we obtain

A

B

x y

AB

uB AB

vB

AB

uA AB

vA

AB A AB

A Θ

θ ,

AB

VB

AB

UB AB

VA

AB

UB

φ

AB B AB

B Θ

θ , φ

− φ

= _A^ABcos ^AB_A sin

AB

A u v

U

AB A AB A = θ Θ

φ + φ

= _A^ABsin ^AB_A cos

AB

A u v

V

U_B^AB= u_B^ABcosφ−v_B^ABsinφ

AB B AB B = θ Θ

φ + φ

= _B^ABsin _B^ABcos

AB

B u v

V IV

V VI

VII VIII IX

^{AB AB}

A AB

B −Θ = A Θ

U_B^AB−U_A^AB+Θ^AB_AL_ABsinφ=−A^ABx_Bsinφ

− −Θ cosφ= Bcosφ

AB AB

AB A AB A AB

B V L x

V A

ƒ Summary of Moment Area Equations in Global Coordinate System

Remarks:

9 Above equations are limited to a straight segment

9 The displacement is assumed to be continuous everywhere within the segment, e.g.

“shear release” and “axial release” cannot be present within the segment AB 9 The rotation is assumed to be continuous everywhere within the segment, e.g.

“moment release” or “hinge” cannot be present within the segment AB

9 Six kinematical quantities involve in above three equations, i.e. {UA^AB,UB^AB,VA^AB,VB^AB,

AB

ΘA ,Θ^ABB }, three for end A {UA^AB,VA^AB,Θ^ABA } and the other for end B {UB^AB,VB^AB, ΘB^AB}

φ

= φ Θ

−

− _A^{AB AB}_{A AB}cos ^AB _Bcos

AB

B V L x

V A

AB AB A AB

B −Θ = A Θ

φ

−

= φ Θ +

− _A^{AB AB}_{A AB}sin ^AB _Bsin

AB

B U L x

U A

LAB

A

B X Y

x B

x A

AB

UB

M/EI diagram Centroid

AB

VB AB

ΘB

AB

UA AB

VA

AB

ΘA

φ

1^st Moment Area Equation

2^nd Moment Area Equation

3^rd Moment Area Equation

¾

Construction of Conjugate Structure Analogy

Let introduce a fictitious segment which is identical to above segment AB but is subjected to the following system of fictitious forces and moments:

9 Moment equal to UA^ABapplied at point A in positive X-direction 9 Moment equal to V_A^ABapplied at point A in positive Y-direction 9 Force equal to Θ^ABA applied at point A in positive Z-direction 9 Moment equal to U_B^ABapplied at point B in negative X-direction 9 Moment equal to VB^ABapplied at point B in negative Y-direction 9 Force equal to Θ_B^ABapplied at point B in negative Z-direction

9 Force equal to A^ABapplied at C.G. of the M/EI diagram in positive Z-direction The free body diagram of the fictitious segment AB showing all fictitious forces and moments is indicated in the figure below.

For the segment AB to be in equilibrium, all fictitious forces and moments must satisfy the following three static equilibrium equations

LAB

A

B

X Y

x B

x A AB

UB

AB

VB AB

ΘB

AB

UA AB

VA AB

ΘA φ

AAB

ΣFZ = 0 Θ^AB_A +A^AB−Θ^AB_B =0

ΣMBX = 0 U_A^AB−Θ^AB_AL_ABsinφ−A^ABx_Bsinφ−U_B^AB=₀

ΣMBY = 0 _{+Θ cosφ+ cosφ−} ^AB₌₀

B B AB AB

AB A AB

A L x V

V A

By comparing the three static equilibrium equations established on the fictitious segment with the moment area equations of the actual segment, we can deduce the following correspondence:

This correspondence is known as the “Conjugate Structure Analogy” and the fictitious segment subjected to a system of fictitious forces and moments is known as the

“Conjugate Structure”. This conjugate structure analogy provides a useful, convenient, and systematic means to set up three equations relating the displacements and rotations at both ends of the actual segment to the bending moment within the segment without geometric consideration (no need to sketch the elastic or deformed curve). The key step of the analogy is to establish the correct “conjugate structure” and the remaining task only involves setting up static equilibrium equations.

1^st Moment Area Equation ΣFZ = 0

2^nd Moment Area Equation ΣMBX = 0

3^rd Moment Area Equation ΣMBY = 0

Conjugate Structure Analogy for Non-straight Segment

¾

Moment Area Equations in Global Coordinate System

ƒ Consider a non-straight segment which is a part of the structure as shown in the figure below. The segment, denoted by A1A2A3A4, is assumed to consist of three straight segments, i.e. segment A1A2, segment A2A3, and segment A3A4, connecting at point A2 and A3; lengths of segments A1A2, A2A3, and A3A4 are denoted by L12, L23, and L34, respectively.

In addition, the segment A1A2A3A4 contains a “moment release” or “hinge” at point B.

ƒ Local coordinate system: The orientation of all segments is indicated by their local coordinate system. The local coordinate system of the segment AiAi+1 is defined such that its origin is at point Ai, x-axis directs along the member, y-axis directs perpendicular to the member, and z-axis points outward of the paper. The orientation of the segment AiAi+1

relative to the global coordinate system is denoted by an angle φi that is measured from the global X-axis to the local x-axis; φi is positive if it directs counterclockwise (for example, in above figure, φ1 and φ2 are positive and φ3 is negative)

ƒ Bending moment and M/EI diagram: Sign convention of the bending moment M at any cross section of each segment follows the standard convention for bending moment of a beam with the local coordinate system defined above. The M/EI diagram is simply obtained by dividing the BMD by the flexural rigidity EI. The distances from the centroid of M/EI diagram of the segment AiAi+1 to the end points Ai and Ai are denoted by x_iⁱⁱ⁺¹andx_iⁱ₊ⁱ₁⁺¹, respectively. The M/EI of the segment A1A2A3A4 is indicated in the figure below.

L12

A1 X

Y

L23

L34

A2

A3

A4

x

x

y x

y

y

φ1

φ2

φ3

Positive moment x y

x y

Negative moment

ƒ Displacements and Rotations: The X-component and Y-component of the displacement and the rotation of point Ai of the segment AiAi+1 are denoted by Uⁱiⁱ⁺¹,Viⁱⁱ⁺¹ and ⁱiⁱ¹

Θ +, respectively, and the X-component and Y-component of the displacement and the rotation of point Ai+1 of the segment AiAi+1 are denoted by Uiⁱ+ⁱ1⁺¹,Viⁱ+ⁱ1⁺¹ and ⁱiⁱ1¹

+

Θ+ , respectively.

C.G.

12

x 1 12

x 2 23

x2

23

x3

34

x3 34

x4

C.G.

C.G.

A1

A2

A3

A4

X Y

12

U1 12

V1

12

Θ1

23 2 12

2 U

U ,

23 2 12

2 V

V ,

12

Θ2

23

Θ2

34 3 23

3 U

U ,

34 3 23

3 V

V ,

23

Θ3 Θ³⁴₃

34

V4 ₃₄

Θ4

34

U4

A1

A2

A3

A4

ƒ Moment Area Equations for Segment A1A2

ƒ Moment Area Equations for Segment A2A3

ƒ Moment Area Equations for Segment A3A4

ƒ Continuity at Point A2

ƒ Continuity at Point A3

1 A A

1 cos

cosφ = ^{1 2} φ

Θ

−

− 12 ¹²2

12 1 12 1 12

2 V L x

V A

2 1A

AA 12

1 12 2 −Θ = Θ

1 A A

1 sin

sinφ =− ^{1 2} φ

Θ +

− ₁^{12 12}_{1 12} ¹²₂

12

2 U L x

U A

2 A A

2 cos

cosφ = ^{2 3} φ

Θ

−

− ₂^{23 23}_{2 23 3}²³

23

3 V L x

V A

3 2A

AA 23

2 23

3 −Θ =

Θ

2 A A

2 sin

sinφ =− ^{2 3} φ

Θ +

− 23 3²³

23 2 23 2 23

3 U L x

U A

3 A A

3 cos

cosφ = ^{3 4} φ

Θ

−

− ₃^{34 34}_{3 34 4}³⁴

34

4 V L x

V A

4 3A

AA 34

3 34

4 −Θ =

Θ

3 A A

3 sin

sinφ =− ^{3 4} φ

Θ +

− 34 4³⁴

34 3 34 3 34

4 U L x

U A

²³

2 12

2 U

U =

23 2 12

2 V

V =

34 3 23

3 U

U =

34 3 23

3 V

V =

34 3 23 3 =Θ Θ

ƒ Relative Rotation at hinge point A2

ƒ Three Moment Area Equations for Segment A1A2A3A4

9 Combine moment area equations of three segments and then apply the continuity conditions and the relative rotation at the hinge point to eliminate displacements and rotations at points A2 and A3

9 Unknowns appearing in the equations are only in terms of displacements and rotations at the ends of the segment and in terms of relative rotation at the hinge point. For this particular case, there are 3 unknowns at point A1, 3 unknowns at point A4, and 1 unknown at point A2 due to presence of a hinge.

ψ2

= Θ

− Θ ¹²2

23 2

4 3 3 2 2

1A AA AA

A

2=A +A +A

ψ

− Θ

− Θ³⁴₄ ¹²₁

[ ]

[ ]

[ ]

[ ]

[

3

]

A A

3 2

A A

3 2

1 A A

3 2

1

3 2

2

sin

sin sin

sin sin

sin

sin sin

sin

sin sin

4 3

3 2

2 1

φ

−

φ + φ

−

φ + φ + φ

−

=

φ + φ + φ Θ +

φ + φ ψ +

−

34 4

34 23

3

34 23

12 2

34 23

12 12 1

34 23

12 1 34 4

x

L x

L L

x

L L

L

L L

U U

A A A

[ ]

[ ]

[ ]

[ ]

[

3

]

A A

3 2

A A

3 2

1 A A

3 2

1

3 2

2

cos

cos cos

cos cos

cos

cos cos

cos

cos cos

4 3

3 2

2 1

φ +

φ + φ +

φ + φ + φ

=

φ + φ + φ Θ

−

φ + φ ψ

−

−

34 4

34 23

3

34 23

12 2

34 23

12 12 1

34 23

12 1 34 4

x

L x

L L

x

L L

L

L L

V V

A A A

1^st Moment Area Equation

2^nd Moment Area Equation

3^rd Moment Area Equation

¾

Construction of Conjugate Structure Analogy

Let introduce a fictitious segment which is identical to the segment A1A2A3A4 but is subjected to the following system of fictitious forces and moments:

9 Moment equal to U₁¹²applied at point A1 in positive X-direction 9 Moment equal to V1¹²applied at point A1 in positive Y-direction 9 Force equal to Θ¹²₁applied at point A1 in positive Z-direction 9 Moment equal to U4³⁴applied at point A4 in negative X-direction 9 Moment equal to V₄³⁴applied at point A4 in negative Y-direction 9 Force equal to Θ³⁴4 applied at point A4 in negative Z-direction 9 Force equal to ψ₂applied at point A2 in positive Z-direction 9 Force equal to A^A1A2applied at C.G. of the M/EI diagram of the

segment A1A2 in positive Z-direction

9 Force equal to A^A2A3applied at C.G. of the M/EI diagram of the segment A2A3 in positive Z-direction

9 Force equal to A^A3A4applied at C.G. of the M/EI diagram of the segment A1A2 in positive Z-direction

The free body diagram of the fictitious segment A1A2A3A4 showing all fictitious forces and moments is indicated in the figure below.

X

2 Y

1A

AA

A1

A2

A3

A4 12

U1

12

V1 12

Θ1

34

V4 34

Θ4 34

U4 12

x 1 12

x 2

23

x2

23

x3

34

x3

34

x4 3

2A

AA

4 3A

AA

ψ2

φ1

φ2

φ3 L12

L23

L34

For the segment A1A2A3A4 to be in equilibrium, all fictitious forces and moments must satisfy the following three static equilibrium equations

It is evident that these three static equilibrium equations established on the fictitious segment are identical to the three moment area equations of the actual segment; i.e. the following correspondence can be deduced:

ΣFZ = 0

0 M_AX

4 =

Σ

0

4 3 3 2 2

1A AA AA

A

2+ + + =

ψ + Θ

−

Θ ³⁴4 A A A

12 1

[ ]

[ ]

[ ]

[ ]

[

^sinsin

]

^{0 sin}

sin sin

sin

sin sin

sin

sin sin

3 A A

3 2

A A

3 2

1 A A

3 2

1

3 2

2

4 3

3 2

2 1

= φ

−

φ + φ

−

φ + φ + φ

−

φ + φ + φ Θ

−

φ + φ ψ

−

−

34 4

34 23

3

34 23

12 2

34 23

12 12 1

34 23

34 4 12 1

x

L x

L L

x

L L

L

L L

U U

A A A

[ ]

[ ]

[ ]

[ ]

[

^coscos

]

^{0 cos}

cos cos

cos

cos cos

cos

cos cos

3 A

A

3 2

A A

3 2

1 A A

3 2

1

3 2

2

4 3

3 2

2 1

= φ +

φ + φ +

φ + φ + φ +

φ + φ + φ Θ +

φ + φ ψ +

−

34 4

34 23

3

34 23

12 2

34 23

12 12 1

34 23

34 4 12 1

x

L x

L L

x

L L

L

L L

V V

A A A

1^st Moment Area Equation ΣFZ = 0

2^nd Moment Area Equation ΣMBX = 0

3^rd Moment Area Equation ΣMBY = 0

0 M_AY

4 =

Σ

Again, the “Conjugate Structure Analogy” can also be established for the non-straight segment in the same fashion as that for the straight segment. This crucial feature renders the method far superior than the method of moment area equations; in particular, it removes several restrictions such as

9 No need to sketch qualitative elastic curve

9 A segment consisting of several straight sub-segments is allowed 9 The method allows the presence of “hinge” within the segment

It is important to remark that there is one restriction on selection of the segment, i.e. the sub-segments constituting the segment must be connected in a series (There is only one starting point and only one end point).

Applications of Conjugate Structure Analogy to Deformation Analysis

¾

Useful Observations

ƒ Unknowns per segment

9 2 components of displacement + 1 rotation at the starting point of segment 9 2 components of displacement + 1 rotation at the end point of segment 9 Relative rotation at hinge points

ƒ Number of equations per segment

9 1 equilibrium of forces in Z-direction; ΣF_Z=0 9 1 equilibrium of moments in X-direction; ΣM_X=0 9 1 equilibrium of moments in Y-direction;ΣMY=0

ƒ Prescribed displacements or rotations at support 9 Roller support Æ 1 component of displacement is zero

9 Pinned or hinge support Æ 2 components of displacement are zero 9 Fixed support Æ 2 components of displacement & rotation are zero 9 Guide support Æ 1 component of displacement & rotation are zero 9 Nonrotational support Æ rotation is zero

Roller support Pinned support

Fixed support Guide support Nonrotational support

¾

Procedure for Deformation Analysis of Determinate Structures 1.

Define the coordinate systems

9 Global coordinate system 9 Local coordinate systems

2.

Determine support reactions and draw BMD

9 Solve for support reactions using static equilibrium 9 Draw BMD based on local coordinate defined in step

1 3.

Choose a proper (straight or non-straight) segment

9 Choose segment containing three unknown displacements and rotations if possible 9 Define the starting point and the end point

9 Obtain local coordinate system for all sub-segments

4.

Construct M/EI diagram for the selected segment

9 Identify flexural rigidity EI of the selected segment

9 Use BMD from step

2

and local coordinate systems from step

3

5.

Construct the conjugate structure of the selected segment

9 At a point on the selected segment where the centroid of M/EI diagram is located,

ƒ Force equal to the area of M/EI diagram is applied in the positive Z-direction 9 At a hinge point (if exists),

ƒ Force equal to the rotation of the hinge is applied in the positive Z-direction 9 At the starting point of the selected segment,

ƒ Force equal to rotation at the starting point is applied in positive Z-direction

ƒ Moment equal to displacement in X-direction of the starting point is applied in positive X-direction

ƒ Moment equal to displacement in Y-direction of the starting point is applied in positive Y-direction

9 At the end point of the selected segment,

ƒ Force equal to rotation at the end point is applied in negative Z-direction

ƒ Moment equal to displacement in X-direction of the end point is applied in negative X-direction

ƒ Moment equal to displacement in Y-direction of the end point is applied in negative Y-direction

6.

Establish 3 equilibrium equations (i.e. ΣFZ=0, ΣMX=0,ΣMY=0)

7.

Solve three equations obtained from step

6

if they contain only three unknowns.

Otherwise, repeat steps

3, 4, 5,

and

6

until the number of equations are sufficient for solving all unknowns

Example 1 Given a statically determinate frame subjected to a concentrated load P as shown in the figure. The flexural rigidity of the member AB, member BC, member CD, and member CE are assumed to be 2EI, 2EI, EI, and EI, respectively. Determine the displacements and rotations at points D and E.

Solution

¾ Define the global coordinate system X-Y and local coordinate systems x-y for member AB, member BC, and member DCE as shown below

L

A B

D

L

L C

P

L E

X,x Y,y

A B

D C E x

x

y y

¾ Compute the support reactions and then draw BMD

¾ The presence of a fixed support at point A indicates that

Since the displacement and rotation at point A are already known, any segment that contains a point A as its starting point or its end points has only three unknowns.

¾ To compute the displacement and rotation at point D, the segment ABCD is chosen with the starting point denoted by A and the end point denoted by D. The local coordinate systems for the sub-segments AB, BC, and CD follow automatically after the starting and end points are defined.

UA = VA = ΘA = 0 A

B D

C P

E

RAX=0

RAY=P MA=PL

-PL

PL

PL

PL

-PL

x y

A B

D C

x

y x

y

¾ The M/EI diagram of the segment ABCD based on the local coordinate systems for ABCD is shown in the figure below

The areas of the M/EI diagramA1^AB,A2^AB,A^BC, and A^CDare given by

The centroid of each M/EI diagram is shown in the figure above

A B

D C

PL/EI

L/3 L/3

L/2

L/2 L/3

2L/3

2L/3 2L/3

( )(

1/2 PL/2EI

)( )

L PL²/4EI

AB

1 = − =−

A

( )(

1/2 PL/2EI

)( )

L PL²/4EI

AB

2 = =

A

(

PL/EI

)( )

L PL²/EI

BC= =

A

( )(

1/2 PL/EI

)( )

L PL²/2EI

CD= =

A

ACD

PL/EI

PL/EI ABC

PL/2EI

-PL/2EI

AB

A1

AB

A2

¾ The conjugate structure of the segment ABCD is shown below

¾ Set up three equilibrium equations for the conjugate structure and then solve for the displacement and rotation at point D

A

B

D C

L/3 L/3

L/2

L/2 L/3

2L/3

2L/3 2L/3

AB

A1 A^2AB

ABC

ACD

=0

AB

UA

=0

AB

VA

=0 Θ^ABA

CD

VD CD

ΘD CD

UD

Y

X

ΣFZ = 0

=0 + + + + Θ

−

Θ^AB_A ^CD_D A₁^AB A₂^AB A^BC A^CD 0

0−Θ − + + + =

3EI PL EI PL 4EI PL 4EI

PL^{2 2 2 2}

CD D

2EI 3PL²

CD D = Θ

ΣMDX = 0

2EI PL 2EI U PL

3 3 CD

D =− =

(

+

)

−

( )

=0

− Θ

−

−U L L L/2

U_A^AB _D^{CD AB}_A A₁^AB A₂^AB A^BC

( )

0 0

0 ⎟⎟⎠ − =

⎜⎜ ⎞

⎝

⎛− +

−

−

− L/2

EI L PL 4EI PL 4EI U PL

2 2 2 CD

D

¾ Next, the displacement and rotation at point E can be computed using either the segment ABCE or the segment DCE; however, the latter segment requires less computation. Let consider the segment DCE, its starting point and end point are chosen to be points D and E, respectively, and the local coordinate system is shown below. The M/EI diagram based on this local coordinate system is given below.

¾ The conjugate structure of the segment DCE is shown below

ΣMDY = 0

3EI 5PL 3EI V 5PL

3 3 CD

D =− =

( ) ⎟=0

⎠

⎜ ⎞

⎝

− ⎛

⎟−

⎠

⎜ ⎞

⎝

− ⎛

⎟⎠

⎜ ⎞

⎝ + ⎛ Θ +

− 3

L 2L 3 2L 3

L 2L V

V_A^AB _D^{CD AB}_A A₁^AB A₂^AB A^BC A^BC

( ) 0

0

0 ⎟=

⎠

⎜ ⎞

⎝

− ⎛

⎟−

⎠

⎜ ⎞

⎝

− ⎛

⎟⎠

⎜ ⎞

⎝

− ⎛ +

− 3

2L 2EI L PL EI PL 3 2L 4EI PL 3 2L 4EI V PL

2 2 2 2

CD D

D C

y

E x

L/3

2L/3 L

( )(

1/2 PL/EI

)( )

L PL²/2EI

DE= − =−

A

ADE

DE

UD DE

VD

DE

ΘD

DE

VE DE

ΘE DE

UE

Y

X D

E ADE

-PL/EI

¾ Set up three equilibrium equations for the conjugate structure and then solve for the displacement and rotation at point E

ΣFZ = 0

=0 + Θ

−

Θ^DE_D ^DE_E A^DE

=0

− Θ

− 2EI

PL 2EI

3PL DE ²

E 2

EI PL²

DE E = Θ

ΣMEX = 0

2EI PL 2EI U PL

3 3 DE

E =− =

=0

− _E^DE

DE

D U

U

=0

−

− ³ U_E^DE 2EI PL

ΣMEY = 0

6EI 5PL 6EI V 5PL

3 3 DE

E =− =

=0

⎟⎠

⎜ ⎞

⎝ + ⎛ Θ +

− 3

L 4L V

V ^DED ^DE

DE E DE

D A

( )

⎟=0

⎠

⎜ ⎞

⎝

− ⎛ +

−

− 3

4L 2EI L PL 2EI V 3PL 3EI

5PL DE ^{2 2}

E 3

Example 2 Given a statically determinate beam subjected to loads as shown in the figure below. The flexural rigidity of the entire beam is assumed to be constant EI. Determine the displacements and rotations at points F and hinge rotation.

Solution

¾ For beam structures, the global and local coordinate structures are defined identically as shown in the figure below

¾ Compute the support reactions and then draw BMD L

B D

L L

qL

L L

A C E

F q

B D

A C E

F X,x Y,y

L

B D

L L

2qL

L L

A C E

F q

RA=qL/2 RC=3qL RE=qL/2

qL²

-2qL²

¾ The presence of supports at point A, C, and E indicates that

It is clear that there is only one segment, the segment CDE, that contains only three kinematical unknowns.

¾ Let consider the segment CDE. The starting point and end point are denoted by points C and E, respectively, and the local coordinate system is the same as those defined above. The M/EI diagram and the conjugate structure of the segment CDE are shown below.

VA = UC = VC = VE = 0

D L L

C E

qL²/EI

-2qL²/EI

x y

CE

A1 CE

A2

2L/3

L/3

CE

A1 CE

UC

CE

VC

CE

ΘC

CE

VE CE

ΘE CE

UE

Y

X

CE

A2

( ) ( ) ( )

EI 2L qL /EI qL 1/2

3 2

CE

1 = =

A

( ) ( ) ( )

EI L qL /EI 2qL - 1/2

3 2

CE

2 = =−

A

C

E

¾ Set up three equilibrium equations for the conjugate structure and then solve for the displacement and rotation at point E

¾ Now, the displacement and rotation at point C and E are already known. The displacement and rotation at point F can then be readily computed using either the segment CF or the segment EF; however, the latter segment requires less computation. Let consider the segment EF with E and F being the starting point and end point, respectively. The local coordinate system of this segment is again the same as those defined above, and the M/EI diagram and the conjugate structure of the segment EF are shown below.

ΣFZ = 0

=0 + + Θ

−

Θ 2^CE

CE 1 CE E CE

C A A

=0

− + Θ

− EI

qL EI qL 6EI

qL CE ^{3 3}

E 3

6EI qL³

CE E = Θ

ΣMEX = 0

=0

CE

UE

=0

− _E^CE

CE

C U

U 0 0−UE^DE =

ΣMEY = 0

6EI qL³

CE C = Θ

( ) ⎟=0

⎠

⎜ ⎞

⎝ + ⎛

⎟⎠

⎜ ⎞

⎝ + ⎛ Θ +

− 3

5L 3 2L 4L V

V 2^CE

CE 1 CE C CE E CE

C A A

( ) 0

0

0 ⎟=

⎠

⎜ ⎞

⎝

− ⎛

⎟⎠

⎜ ⎞

⎝ + ⎛ Θ +

− 3

5L EI qL 3 4L EI 2L qL

3 3 CE C

¾ Set up three equilibrium equations for the conjugate structure and then solve for the displacement and rotation at point E

L

E F x

y

EF

UE

EF

VE

EF

ΘE

EF

VF

EF

ΘF EF

UF

Y

X M/EI ≡ 0

E

F

ΣFZ = 0

=0 Θ

− Θ^EF_E ^EF_F

=0 Θ

− ^EFF 3

6EI qL

6EI qL³

EF F = Θ

ΣMFX = 0

0 0−U_F^EF=

=0

− _F^EF

EF

E U

U

=0

EF

UF

ΣMFY = 0 V −V +Θ^EFE

( )

L =⁰

EF F EF E

( )

⁰

0− + L =

6EI V qL

3 EF F

6EI V qL

4 EF F =

¾ Next, rotation of the hinge at point B can be computed using either the segment AC.

Let the starting point and the end point be denoted by A and C, respectively. The M/EI diagram and the conjugate structure of the segment AC are shown below

¾ The rotation at the hinge can be obtained from moment equilibrium about Y-axis at point A:

B L L

A C

qL²/EI x y

AC

A1

2L/3

L/2

AC

A1 AC

UA

AC

VA AC

ΘA

AC

VC AC

ΘC

AC

UC Y

X

AC

A2

( ) ( ) ( )

EI 2L qL /EI qL 1/2

3 2

AC

1 = =

A

( ) ( ) ( )

3EI 2L 4qL /EI 2qL - 1/3

3 2

AC

2 = =−

A

A

C

AC

A2

-2qL²/EI

ψB

ΣMAY = 0 ( ) ⎟=0

⎠

⎜ ⎞

⎝

− ⎛

⎟⎠

⎜ ⎞

⎝

− ⎛ ψ

− Θ +

− 2

3L 3

L 4L 2L V

V_A^AC _C^AC _C^AC _B A₁^AC A₂^AC

EI qL³

B= ψ

( ) 0

0

0 ⎟=

⎠

⎜ ⎞

⎝ + ⎛

⎟⎠

⎜ ⎞

⎝

− ⎛ ψ

− +

− 2

3L 3EI 4qL 3 4L EI L qL 6EI 2L

qL ^{3 3}

B 3

Example 3 Given a statically determinate frame subjected to external loads as shown in the figure. The flexural rigidity of the member AB, member BD, member DE, and member DF are assumed to be 2EI, EI, 2EI, and EI, respectively. Determine the displacements and rotations at points F and the hinge rotation at point C.

Solution

¾ Define the global coordinate system X-Y and local coordinate systems x-y for member AB, member BDF, and member DE as shown below

A

B D

L C

2qL

E qL

L L/2 L

F

L/2

B C D F

A E

Y

X y

y y x

x x 2q

¾ Compute the support reactions and then draw BMD

¾ The presence of pinned supports at point A and E indicates that

¾ The hinge rotation can be obtained first from the segment ABDE since it contains only three unknowns, i.e.Θ^AB_A,Θ^DE_E,ψ_C. The starting point and end point of the segment are denoted by A and E, respectively, and the local coordinate systems are given below.

UA = VA = UE = VE = 0 A

B C D

2qL

E

qL F

RAX=3qL/2

RAY=qL

2q

RAY=3qL RAX=-5qL/2

qL²/2 -qL²/2

-qL²/2

-qL² -qL^<