chapter_5.11-12_isa.pdf

DDGDDGDDG

Department of Continuum Mechanics and Structural Engineering

Bachelor in Aerospace Engineering

Departamento de Mecánica de Medios Continuos y Teoría de Estructuras

DDGDDGDDG

Introduction to Structural Analysis

Chapter 5. Displacements in beamsSection 11 and 12

Index

INDEX

Chapter 5. Displacements in beams

Section 11 and 12


• Introduction• Relationship between internal forces and displacements• Navier-Bresse´s formulas• Plane beams• Navier-Bresse´s formulas in straight beams• Strain energy density• Energy theorems• Summary• References

STRESS TENSORS


Internal forces/displacements relationships

Zzx

y

y

Beam ( )( ), ,

, ,

x y

x y T

F Q Q N

M M M M

=

=

r

r

Internal forces/displacements Relationships


X, Y, Z: Global coordinate system

x, y, z: Local coordinate system

X

Yx

z

( )( )

, ,

, ,

x y z

x y z

du du du du

d d d dφ φ φ φ

=

=

r

r

x

y

φr

Kinematic hypotheses

( ) ( )( ) ( )

o

x x z

o

y y z

u u z z y

u u z z x

φ

φ

= − ⋅

= + ⋅

ur



z

x

y

z

u

u u

u

=

rx

y

z

φφ φ

φ

=

r

Local coordinate system

( ) ( )( ) ( ) ( )

y y z

o

z z y x

u u z z x

u u z z x z y

φ

φ φ

= + ⋅

= − ⋅ + ⋅

vector ntdisplaceme=ur

vector rotation=φr

x

y

φr

:

Bending:

ur ( ) ( )

( ) ( )

2

2

1

1

x y y xyx

x y xy

y x xy y x

M z I M z Pd

dz E I I P

d M z P M z I

dz E I I P

φ

φ

⋅ + ⋅= ⋅

⋅ −

⋅ + ⋅= ⋅

⋅ −



z ( )

( ) ( )

( ) ( )

o

z

o

xxz y

o

y

yz x

N zdu

dz E A

duz z

dz

duz z

dz

γ φ

γ φ

=⋅

= +

= −

( )TzM zd

dz G J

φ =⋅

Torsion

x y xydz E I I P⋅ −

x

y

φr ( )

( )( ) ( )

( )( )

( ) ( )

2

2

x x

y ex y ey xy

yz

o x y xy

x x

ex xy ey xx

o x y xy

Q z m y I m y P

a y I I P

m y P m y IQ z

a y I I P

τ ⋅ − ⋅

= ⋅ − ⋅ −

⋅ − ⋅− ⋅ ⋅ −

ur

Shear stresses



z

( )

( )( )

( ) ( )

( )( )

( ) ( )

2

2

o x y xy

y y

y ex y ey xy

xz

o x y xy

y y

ex xy ey xx

o x y xy

a y I I P

Q z m x I m x P

b x I I P

m x P m x IQ z

b x I I P

τ

⋅ −

⋅ − ⋅= ⋅ − ⋅ −

⋅ − ⋅− ⋅ ⋅ −

x

yur

φr

Additional hypotheses

• Principal axis

• Double symmetric cross-sections



z( )( )

( )

( )( )

( )

x

y ex

yz

o x

y

eyx

xz

o y

Q z m y

a y I

m xQ z

b x I

τ

τ

= ⋅

= ⋅

( )

( )

1

1

xx

x

y y

y

M zd

dz E I

d M z

dz E I

φ

φ

= ⋅

= ⋅

x

yur

φr

yzτγ =

( )( )

( )

( )( )

( )

x

y ex

yz

o x

y

eyx

xz

o y

Q z m y

a y I

m xQ z

b x I

τ

τ

= ⋅

= ⋅



z

( )

( ) ( )

( ) ( )

o

z

o

xxz y

o

y

yz x

N zdu

dz E A

duz z

dz

duz z

dz

γ φ

γ φ

=⋅

= +

= −

yz

yz

xzxz

G

G

τγ

τγ

=

=

The displacements depend on x, y!

( )( )

( )

( )( )

( )

x

y ex

yz

o x

y

y ex

xz

o y

Q z m y

a y I

Q z m x

b x I

τ

τ

= ⋅

= ⋅

x

yur

φr

Assuming a constant shear stress



z

( )

( )

y

yz

cy

x

xz

cx

Q z

A

Q z

A

τ

τ

=

=

Assuming a constant shear stress

Acx: Shear area due to Qx

Acy: Shear area due to Qy

( )( )

( )

( )( )

( )

x

y ex

yz

o x

yy ex

xz

o y

Q z m y

a y I

Q z m x

b x I

τ

τ

= ⋅

= ⋅

( )

( )

y

yz

cy

x

xz

cx

Q z

A

Q z

A

τ

τ

=

=

Matching the strain energy per unit length due to shear stress



2 2yz xz

L

A

U dAG G

τ τ= + ⋅∫

( ) ( )( )

( ) ( )( )

2

22

1

x

y ex

L

o xA

x

y ex

o xA

Q z m yU dA

G a y I

Q z m ydA

G a y I

⋅= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

∫

∫

Assuming that only applies a shear force Qy

( ) ( )22

1 y y

L

cy cyA

Q z Q zU dA

G A G A

= ⋅ ⋅ = ⋅

∫

( )( )

( )

( )( )

( )

x

y ex

yz

o x

yy ex

xz

o y

Q z m y

a y I

Q z m x

b x I

τ

τ

= ⋅

= ⋅

( )

( )

y

yz

cy

x

xz

cx

Q z

A

Q z

A

τ

τ

=

=


Matching the strain energy per unit length due to shear stress


( )( )

cy 2

1 A

x

ex

o xA

m ydA

a y I

=

⋅ ⋅ ∫

( )( )

cx 2

1 A

y

ey

o yA

m xdA

b x I

=

⋅ ⋅ ∫ c

5 A

6A= ⋅

c

9 A

10A= ⋅ c

30 A

31A= ⋅

STRESS TENSORS


Navier-Bresse´s formulas

Zzx

y

y

( )( ), ,

, ,

x y

x y T

F Q Q N

M M M M

=

=

r

r

A

B



X

Yx

z

( )( )

, ,

, ,

x y z

x y z

du du du du

d d d dφ φ φ φ

=

=

r

r

A

Integrating with respect to a

reference section

( ) ( )( )( ) ( )( )

, , ,

, , ,

B A A

B A A

u f u M A B F A B

u M A B F A B

φ

φ φ

= − −

= − −

r r rr r

r r r rr

X, Y, Z: Global coordinate system

x, y, z: Local coordinate system

Aφ

The rotations also produce displacements!

In a plane problem



B A ABu dφ= ⋅

A BB A ABu rφ= ∧

r rv

Small displacement

• In a 3D beam

Z

B

z

xy

Z

B

z

xy

Z

B

z

xy

Infinitesimal displacement vector in a slice

expressed in global axes X,Y,Z.=ud r

B

B AA

B B

B A A ABA A

d

u u r du d r

φ φ φ

φ φ

= +

= + ∧ + + ∧

∫

∫ ∫

r r r

r rr r rv v

Plane Beams


X

Y

Z

A

z

P

X

Y

Z

A

z

X

Y

Z z

Pexpressed in global axes X,Y,Z.

Infinitesimal rotation vector in a slice

expressed in global axes X, Y, Z.

Vector P-B in global axes X, Y, Z.

Vector P-B in global axes X, Y, Z.

=θrd

=rr

=ABrr

STRESS TENSORS


Plane beams

L

A B

q

Z A

By

z

PZ

P

Plane Beams


straight beams

YY

Curve beams

Arcs

q Qy , Mx

A transversal load does not produce axial force

Y

BBBy

z

P

( )

( ) ( )

o

z

oxx

N zdu

dz E A

Q zduzφ

=⋅

= +

( )

( )

xx

x

M zd

dz E I

d M z

φ

φ

=⋅

Plane Beams

MovementIn local axis


Z

A

( )( )0, ,

,0,0

y

x

F Q N

M M

=

=

r

r

( ) ( )

( ) ( )

xxy

cx

o

y y

x

cy

Q zduz

dz G A

du Q zz

dz G A

φ

φ

= +⋅

= −⋅

( )y y

y

z T

d M z

dz E I

d M

dz G J

φ

φ

= −⋅

=⋅

( )

( )

,0,0 ,0,0

0, , 0, ,

xx

x

y

y z

cy

M dsd d

E I

Q Ndu du du

G A E A

φ φ ⋅= = ⋅

= = ⋅ ⋅

r

r

Internal forcesIn local axis

MovementIn local axis

In global axis

Y

BBBy

z ( )

( )

,0,0 ,0,0

0, , 0, ,

xx

x

y

y z

cy

M dsd d

E I

Q Ndu du du

G A E A

φ φ ⋅= = ⋅

= = ⋅ ⋅

r

r

P

α

dy

Plane Beams


Internal forcesIn local axis

In global axis

Z

A

( )

( )

0, ,

0, ,

B B

B A B A

r y y z z

y y z z

PB

AB

→

→

= = − −

= − −

r

( )( )0, ,

,0,0

y

x

F Q N

M M

=

=

r

r

( )

( )

,0,0 ,0,0

0, , 0, ,

xx

x

y y

y z

cy cy

M dsd d

E I

Q QN Ndu du du dy dz dz dy

G A E A G A E A

φ φ ⋅= = ⋅

= = − + + ⋅ ⋅ ⋅ ⋅

r

r

tandy

dzα =

cosdz ds

dy sen ds

αα

= ⋅= ⋅

Z

BBBy

z

P

Plane Beams


Y

A

( ) ( )

( ) ( )

BB A xx x

Ax

B ByB A A xy y x B A B

A Acy x

B ByB A A xz z x B A B

A Acy x

Mds

E I

Q MNu u z z dz dy z z ds

E A G A E I

Q MNu u y y dy dz y y ds

E A G A E I

φ φ

φ

φ

= − ⋅⋅

= − ⋅ − + ⋅ − ⋅ − ⋅ − ⋅ ⋅ ⋅ ⋅

= + ⋅ − + + + − ⋅ ⋅ ⋅ ⋅

∫

∫ ∫

∫ ∫

STRESS TENSORS


Navier-Bresse´s formulas in straight beams

0yy

0dy

dzds

==

=

= z

y

w u

v u

=

=θ

v

w

y

x

y

M M

Q Q

=

=



0yy BA ==

( ) ( )

B

B AA

B B

B A A B A BA A

c

B

B AA

Mdz

E I

Q Mv v z z dz z z dz

G A E I

Nw w dz

E A

θ θ

θ

= +⋅

= − ⋅ − + ⋅ − ⋅ − ⋅⋅ ⋅

= + ⋅⋅

∫

∫ ∫

∫

xθ φ=

Q N M

zA B

θv

w

y

0yy

0dy

dzds

==

=

= z

y

w u

v u

=

=x

y

M M

Q Q

=

=



( ) ( )

B

B AA

B B

B A A B A BA A

c

B

B AA

Mdz

E I


G A E I

Nw w dz

E A

θ θ

θ

= −⋅

= + ⋅ − + ⋅ − ⋅ − ⋅⋅ ⋅

= + ⋅⋅

∫

∫ ∫

∫

Q N M

z0yy BA ==

xθ φ=

w

θy

v

0yy

0dy

dzds

==

=

= z

y

w u

v u

=

=x

y

M M

Q Q

=

=



( ) ( )

B

B AA

B B

B A A B A BA A

c

B

B AA

Mdz

E I


G A E I

Nw w dz

E A

θ θ

θ

= −⋅

= − ⋅ − − ⋅ + ⋅ − ⋅⋅ ⋅

= + ⋅⋅

∫

∫ ∫

∫

Q N M

z0yy BA ==

xθ φ=

v

w

θy

0yy

0dy

dzds

==

=

= z

y

w u

v u

=

=x

y

M M

Q Q

=

=



( ) ( )

B

B AA

B B

B A A B A BA A

c

B

B AA

Mdz

E I


G A E I

Nw w dz

E A

θ θ

θ

= +⋅

= − ⋅ − − ⋅ − ⋅ − ⋅⋅ ⋅

= + ⋅⋅

∫

∫ ∫

∫

Q N M

z0yy BA ==

xθ φ=

Example:

A B

P

Shear forceQ

Bendingmoment M

A B

P.L

Axial force N

0N =

z

y



A B A B

z

Q P= ( )M P L z= ⋅ − A Bw w=

Boundaryconditions

0A Av θ= =

( ) ( )0 0

L L

B

c

P L zPv dz L z dz

G A E I

⋅ −= ⋅ + ⋅ − ⋅

⋅ ⋅∫ ∫3

3B

c

P L P Lv

G A E I

⋅ ⋅= +⋅ ⋅ ⋅

z

L

Due to bending momentDue to shear force

( ) 2

0 2

B L

BA

P L zM P Ldz dz

E I E I E Iθ

⋅ − ⋅= ⋅ = ⋅ =⋅ ⋅ ⋅ ⋅∫ ∫

Example:

3

3B

c

P L P Lv

G A E I

⋅ ⋅= +⋅ ⋅ ⋅

z

y

tancor te

B

c

P Lv

G A

⋅=⋅

3

3

flexión

B

P Lv

E I

⋅=⋅ ⋅



z

L

c

h

x

y

( )( )

tan

3 2

32

2

3

3

13

12 0,6 1

2 1 1,2

cor te

cB

flexión

B c

P LG Av E I

P Lv G A LE I

E c hh

E c h LL

ν

ν

⋅⋅ ⋅ ⋅= =

⋅ ⋅ ⋅⋅ ⋅

⋅ ⋅ ⋅ ⋅ = = ⋅ ⋅ + ⋅ ⋅ ⋅

⋅ +

The influence of Q in the displacement of Bis much less than the influence of M

Influence of the shear force in the displacement


%bending

shearV

v


νννν = 0.3

Straight beam neglecting the deformations induced by shear and axial force

Navier-Bresse´s formula are:

θ

y



Q N M

θv

w

z

( ) ( )

B

B AA

B

B A A B A BA

B A

Mdz

E I

Mv v z z z z dz

E I

w w

θ θ

θ

= +⋅

= + ⋅ − + ⋅ − ⋅⋅

=

∫

∫

Example:

z

yq

2

2

q L⋅

( )22

q L z⋅ −

A B



( )2 3

0 2 6

B L

BA

q L zM q Ldz dz

E I E I E Iθ

⋅ − ⋅= ⋅ = ⋅ =⋅ ⋅ ⋅ ⋅ ⋅∫ ∫

0A Bw w= =

z

L A B

( ) ( )3 4

0 2 8

B L

BA

q L zM q Lv L z dz dz

E I E I E I

⋅ − ⋅= ⋅ − ⋅ = ⋅ =⋅ ⋅ ⋅ ⋅ ⋅∫ ∫

A B

Mohr’s Theorems The Moment Area Method


First theorem

The change in slope over any length of a member subjected to bending is equal to thearea of the bending diagram over that length divided by the flexural strength (EI)


Diagram of bending moment

A’

B’

Undeformedmid-line

θΒ

θΑ

A B

A’

B’Deformed mid-line θΒ− θΑ



First theorem

The change in slope over any length of a member subjected to bending is equal to thearea of the bending diagram over that length divided by the flexural stiffness (EI)


( )1 B

B AAM z dz

E Iθ θ− = ⋅ ⋅

⋅ ∫

This is one of the Navier-Bresse equations

( ),B A

A M A B

E Iθ θ

−− =

⋅



Second theorem

For an originally straight beam, subject to bending moment, the vertical interceptbetween one terminal and the tangent to the curve of another terminal is the firstmoment of the bending moment diagram about the terminal where the intercept ismeasured, divided by the flexural stiffness (EI)


Diagram of bending moment

A’

B’

Undeformedmid-line

θΑ

A B

A’

B’Deformed mid-line

Av

AAB θ⋅

d

Bv



Second theorem

For an originally straight beam, subject to bending moment, the vertical interceptbetween one terminal and the tangent to the curve of another terminal is the firstmoment of the bending moment diagram about the terminal where the intercept ismeasured, divided by the flexural stiffness (EI)


( ) ( )( ),e

B A A

m A M Bd v v AB

E Iθ= − + ⋅ =

⋅

( ) ( )1

B

B A A BA

d v v AB M z z dzE I

θ= − + ⋅ = ⋅ − ⋅⋅ ∫

This is one of the Navier-Bresse equations

Bending moment

P·L

Example:

z

y

A B


Mohr’s Theorems


A B

z

L

( ), 1 1

2B A

A M A BL P L

E I E Iθ θ

−− = = ⋅ ⋅ ⋅ ⋅

⋅ ⋅

( ) ( )( ), 1 1 2

2 3

e

B A A

m A M Bd v v AB L P L L

E I E Iθ= − + ⋅ = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅

A B

1

2ij ij

D D

V U dV dVσ ε= ⋅ = ⋅ ⋅ ⋅∫ ∫z

y

iX

it

tD∂

uD∂

D

iu

Strain Energy Density


xy i

X

Z

Y

x

z

y

2 22 2 221

2

y yx x T

x y cy cxL

M QM Q MNV ds

E A E I E I G A G A G J

= ⋅ + + + + + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

∫

The strain energy density in a straight beams is:

1

2ij ij

D

V dVσ ε= ⋅ ⋅ ⋅∫



( )1

2z z xz xz yz yz

L A

V ds dAσ ε τ γ τ γ= ⋅ ⋅ + ⋅ + ⋅ ⋅ ⋅∫ ∫

( )1

2z z xz xz yz yz

D

V dVσ ε τ γ τ γ= ⋅ ⋅ + ⋅ + ⋅ ⋅∫

( ) ( )( ) ( )

o

x x z

o

y y z

u u z z y

u u z z x

φ

φ

= − ⋅

= + ⋅

Kinematic hypotheses Equilibrium equations

z

A

y yz

N dA

Q dA

σ

τ

= ⋅

= ⋅

∫

∫

x z

A

y z

M y dA

M x dA

σ

σ

= ⋅ ⋅

= − ⋅ ⋅

∫

∫



oooy yx xz z

x y x y y x T

L

d dud dudu dV N M M Q Q M ds

dz dz dz dz dz dz

φφ φφ φ

= ⋅ + ⋅ + ⋅ + ⋅ − + ⋅ + + ⋅ ⋅ ∫

( ) ( )( ) ( ) ( )

y y z

o

z z y xu u z z x z yφ φ= − ⋅ + ⋅A

x xz

A

Q dAτ= ⋅

∫

∫ ( )y z

A

T yz xz

A

M x y dAτ τ= ⋅ − ⋅ ⋅

∫

∫

( )1

2z z xz xz yz yz

L A

V ds dAσ ε τ γ τ γ= ⋅ ⋅ + ⋅ + ⋅ ⋅ ⋅∫ ∫

( )

( )

xx

x

y y

y

z T

M zd

dz E I

d M z

dz E I

d M

dz G J

φ

φ

φ

=⋅

=⋅

=⋅

( )

( ) ( )

( ) ( )

o

z

oxx

y

cx

o

y y

x

cy

N zdu

dz E A

Q zduz

dz G A

du Q zz

dz G A

φ

φ

=⋅

= +⋅

= −⋅



2 22 2 221

2

y yx x T

x y cy cxD L

M QM Q MNU dV ds


⋅ = ⋅ + + + + + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

∫ ∫

oooy yx xz z

x y x y y x T

L

d dud dudu dV N M M Q Q M ds

dz dz dz dz dz dz

φφ φφ φ

= ⋅ + ⋅ + ⋅ + ⋅ − + ⋅ + + ⋅ ⋅ ∫

Theorem of virtual forces

Energy theorems

The necessary and sufficient condition for a displacement field u is compatible with astrain field D is that, for all the stress field Tψψψψ in equilibrium with a system of loads f ψ v inB and t ψ ΩΩΩΩ in ∂Bt , it holds that internal and external work are equal


c c

i i i i ij ijD D Dt u dS X u dV dVψ ψ ψσ ε

∂⋅ ⋅ + ⋅ ⋅ = ⋅ ⋅∫ ∫ ∫

Hypothesis:

• No se considera alabeo por torsión

( )y y y yx x x x T T

k kx y x yL L

M M Q QM M Q Q M MN Nds q u ds P M

E A E I E I G A G A E J

ψ ψψ ψ ψψψ ψ ψδ φ

⋅ ⋅⋅ ⋅ ⋅⋅ + + + + + ⋅ = ⋅ ⋅ + ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∑∫ ∫

r rr rr r

X

Z

Y

x

z

y

( ) ( )( ) ( )( ) ( ) ( )

o

x x z

o

y y z

o

z z y x

u u z z y

u u z z x

u u z z x z y

φ

φ

φ φ

= − ⋅

= + ⋅

= − ⋅ + ⋅

Kinematic hypotheses Equilibrium equations

z

A

y yz

A

x xz

N dA

Q dA

Q dA

ψ ψ

ψ ψ

ψ ψ

σ

τ

τ

= ⋅

= ⋅

= ⋅

∫

∫

∫ ( )

x z

A

y z

A

T yz xz

M y dA

M x dA

M x y dA

ψ ψ

ψ ψ

ψ ψ ψ

σ

σ

τ τ

= ⋅ ⋅

= − ⋅ ⋅

= ⋅ − ⋅ ⋅

∫

∫

∫

Energy theorems



x zxz

y zxz

zz

u u

z x

u u

z y

u

z

γ

γ

ε

∂ ∂= +∂ ∂∂ ∂= +∂ ∂

∂=∂

( )ij ij z z xz xz yz yz

D L A

dV ds dAψ ψ ψ ψσ ε σ ε τ γ τ γ⋅ ⋅ = ⋅ + ⋅ + ⋅ ⋅ ⋅∫ ∫ ∫

x xz

A

Q dAτ= ⋅∫ ( )T yz xz

A

M x y dAτ τ= ⋅ − ⋅ ⋅∫

oooy yx xz z

ij ij x y x y y x T

D L

d dud dudu ddV N M M Q Q M ds

dz dz dz dz dz dz

ψ ψ ψ ψ ψ ψ ψφφ φσ ε φ φ

⋅ ⋅ = ⋅ + ⋅ + ⋅ + ⋅ − + ⋅ + + ⋅ ⋅ ∫ ∫

( )

( )

xx

x

y y

y

z T

M zd

dz E I

d M z

dz E I

d M

dz G J

φ

φ

φ

=⋅

=⋅

=⋅

( )

( ) ( )

( ) ( )

o

z

oxx

y

cx

o

y y

x

cy

N zdu

dz E A

Q zduz

dz G A

du Q zz

dz G A

φ

φ

=⋅

= +⋅

= −⋅

Energy theorems



y x y y yx x T Tij ij

x y cx cyD L

M M Q Q Q QM M M MN NdV ds


ψ ψ ψψ ψψψσ ε

⋅ ⋅ ⋅⋅ ⋅⋅⋅ ⋅ = + + + + + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∫ ∫

oooy yx xz z

ij ij x y x y y x T

D L

d dud dudu ddV N M M Q Q M ds

dz dz dz dz dz dz

ψ ψ ψ ψ ψ ψ ψφφ φσ ε φ φ

⋅ ⋅ = ⋅ + ⋅ + ⋅ + ⋅ − + ⋅ + + ⋅ ⋅ ∫ ∫

c c

i i i iD D

L

t u dS X u dV q u dsψ ψ ψ∂

⋅ ⋅ + ⋅ ⋅ = ⋅ ⋅∫ ∫ ∫r r ( )

k k

P Mψ ψδ φ+ ⋅ + ⋅∑r rr r

Point load and moments

Energy theorems



( )y y y yx x x x T T

k kx y x yL L

M M Q QM M Q Q M MN Nds q u ds P M

E A E I E I G A G A E J

ψ ψψ ψ ψψψ ψ ψδ φ

⋅ ⋅⋅ ⋅ ⋅⋅ + + + + + ⋅ = ⋅ ⋅ + ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∑∫ ∫

r rr rr r

θv

w

y

Ly yQ QM MN N

ψψψ ⋅⋅⋅ + + ⋅ = ∫

Straight beam

Energy theorems


Application of the theorem to calculate the movements of a be am


Q N M

w

zA B

( ) ( ) ( )0

0

y yx x

x y

L

y y z z y y z z x xk kk k

Q QM MN Ndz

E A E I G A

q u q u dz P P Mψ ψ ψ ψ ψδ δ φ

⋅⋅⋅ + + ⋅ = ⋅ ⋅ ⋅

= ⋅ + ⋅ ⋅ + ⋅ + ⋅ + ⋅

∫

∑ ∑∫

In structures compose by several beams, this equation is applied to each beam andadded

Real state

• The real beam is considered

• A point force (or moment) equal to one isapplied at the point and direction in whichthe displacement (or rotation) want to be

Energy theorems




A B

A B

Real state

Virtual state

P

P

Pψ = 1

0

1

Ly y px x

y

x y

Q QM MN Ndz

E A E I G A

ψψψ

δ ⋅⋅⋅ + + ⋅ = ⋅ ⋅ ⋅ ⋅ ∫

the displacement (or rotation) want to becalculated

Castigliano's theorems

Pi∆∆∆∆i

In each point i where a load P is applied the

First theorem

The derivative of the strain energy of a solid in equilibrium with respect to a generalizedmovement of a point is equal to the generalized force applied to it.

Energy theorems


x

z

y

P1

Pn

∆∆∆∆1

∆∆∆∆n

In each point i where a load Pi is applied thegeneralized movement ∆i is defined in the directionof its line of action

( )N

i i i

i

V PΠ = ∆ − ⋅ ∆∑

Strain energy of the solid as a function of ∆∆∆∆i

Total potential energy of the solid i

Work of the generalized forces

P1

Pi

∆∆∆∆1

∆∆∆∆i

∆∆∆∆n

( )N

i i i

i

V PΠ = ∆ − ⋅∆∑

i0 δ δΠ = ∀ ∆Solid in equilibrium 0i

i i

δ∂Π ∆ =∂∆∑

Theorem of Stationary Potential Energy

Energy theorems



x

z

y

P1

Pn

∆∆∆∆n

i

i

VP

∂ =∂∆

0i i

i i

VP δ

∂ − ⋅ ∆ = ∂∆ ∑

As must be met for any displacement δ∆δ∆δ∆δ∆i

Theorem of Stationary Potential Energy

Using this theorem the generalizedforces acting on the structure can becalculated

Pi∆∆∆∆i

Energy theorems


Second theorem

The derivative of the complementary strain energy of a solid in equilibrium with respectto a generalized force of a point is equal to the generalized displacement.

In each point i where a load P is applied the


x

z

y

P1

Pn

∆∆∆∆1

∆∆∆∆n

Complementary Total potential

energy of the solid

Complementary work of the

generalized forces

( )N

c c

i i i

i

V P PΠ = − ∆ ⋅∑

Complementary Strain energy of the solid as a

function of P i

In each point i where a load Pi is applied thegeneralized movement ∆i is defined in the directionof its line of action

P1

Pi

∆∆∆∆1

∆∆∆∆i

∆∆∆∆n

i0 c Pδ δΠ = ∀ 0c

i

i i

PP

δ∂Π =∂∑

cV ∂∑

( )N

c c

i i i

i

V P PΠ = − ∆ ⋅∑

Energy theorems

Solid in equilibrium

(Theorem of Stationary Complementary potential Energy)



x

z

y

P1

Pn

∆∆∆∆n

c

i

i

V

P

∂ = ∆∂

0c

i i

i i

VP

Pδ

∂ − ∆ ⋅ = ∂ ∑

As must be met for any displacement δδδδPi

In a linear elastic solid V = V c

i

i

V

P

∂ = ∆∂

Theorem of Crotti-Engesseror

1st theorem of EngesserUsing the second theorem thegeneralized displacement of thestructure can be calculated.Castigliano's method usually refers tothe application of this theorem.

State I

Reciprocal work theorem

P1 P2

Energy theorems


A force (or moment) equal to one is applied atthe point and direction in which thedisplacement (or rotation) want to becalculated


A B

A B

State I(Real state)P

P

P = 11I II

i i p

i

P ⋅∆ = ⋅ ∆∑

State II(Virtual State)

M

It is necessary to calculatethe displacement in thevirtual state, of all points inthe state where there areactual applied load

Z

x

y

Strain energy

2 22 2 221 y yx x TM QM Q MN

V ds

= ⋅ + + + + + ⋅ ∫

Summary


XY

z 2

y yx x T

x y cy cxL

V dsE A E I E I G A G A G J

= ⋅ + + + + + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∫


0

1

Ly y px x

y

x y

Q QM MN Ndz

E A E I G A

ψψψ

δ ⋅⋅⋅ + + ⋅ = ⋅ ⋅ ⋅ ⋅ ∫

z

P1

Pi

Pn

∆∆∆∆1

∆∆∆∆i

∆∆∆∆n

1st Theorem of Castigliano

i

i

VP

∂ =∂∆ i

i

V

P

∂ = ∆∂

2sd Theorem of Castigliano

Linear elastic material

Summary


x

z

y c

i

i

V

P

∂ = ∆∂

Theorem of Crotti-Engesser

I II II I

i i i i

i i

P P⋅ ∆ = ⋅∆∑ ∑

Reciprocal work theorem

References

BOOK CHAPTERS

J.A. Garrido, A. Foces, Resistencia de Materiales. Universidad de

Valladolid, 1994.

2 and 9

Ortiz Berrocal, L “Resistencia de Materiales”, Ed. McGraw Hill,

1998

5 and 9

P.P. Benham, R.J. Crawford, Mechanics of Materials, Longman 7


P.P. Benham, R.J. Crawford, Mechanics of Materials, Longman

Scientific and Technical, 1991.

7

Samartin Quiroga, A. “Resistencia de Materiales”, Ed. Bellisco. 1990 2,3 and 4

Given the structure of the figure, determine the horizontal displacement of point C.

The effect of axial and shear in the movements is not considered

8

Example


4

2

A

B C

D

8 m

4 m

4

8B C

4 m

Calculation of reactions

0

0

H

V

F

F

=

=

=

∑∑∑

Example


2

A D

8 m

4 m

6,51,5

5 1

0

0

A

C

M

M

=

=∑∑

B CMoment distribution

Stretch AB

( ) 5M s s= ⋅

s

s

Example


A D

( ) 5M s s= ⋅

Stretch BC

( ) 320

2M s s= + ⋅

( ) 1352

2M s s= − ⋅

0 < s < 4

4 < s < 8

Stretch CD

( )M s s=

( ) 4M s s= −

0 < s < 2

2 < s < 4

s

1

B C

Example


A D

Virtual state

1

B C

( )M s sψ =

Moment distribution

Stretch ABs

s

Example


A D

0,50,5

1

( )M s s=

Stretch BC

( ) 42

sM sψ = −

s

Virtual state

3 13s s

( )3

1 0

1

L

xi xic

i x i

M Mdz u

E I

ψ

=

⋅ ⋅ = ⋅⋅∑ ∫

Example


( ) ( ) ( )4 4 4

0 0 0

3 1320 4 52 4

5 2 2 2 20 1 c

x x xi i i

s ss s

s sds ds ds u

E I E I E I

+ ⋅ ⋅ − − ⋅ ⋅ − ⋅ ⋅ ⋅ + ⋅ + ⋅ + = ⋅⋅ ⋅ ⋅∫ ∫ ∫

382 1

3cu

E I= ⋅

⋅

1 B C

sStretch BC

Moment distribution

Example


A D

1/81/8

( ) 118

M s sψ = − ⋅

To calculate the rotation in B

Virtual state

3 13s s

( )3

1 0

1

L

xi xic

i x i

M Mdz u

E I

ψ

=

⋅ ⋅ = ⋅⋅∑ ∫

Example


( ) ( )4 4

0 0

3 1320 1 52 1

2 8 2 80 0 1 B

x xi i

s ss s

ds dsE I E I

θ

+ ⋅ ⋅ − − ⋅ ⋅ − − ⋅ − ⋅ + = ⋅

⋅ ⋅∫ ∫

128 1

3B

E Iθ = − ⋅

⋅

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