# deformations of statically determinate bar deformations of statically determinate bar structures.

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• Department of Structural Mechanics

Faculty of Civil Engineering, VŠB-Technical University of Ostrava

Statics of Building Structures I., ERASMUS

Deformations of statically

determinate bar structures

• 2 / 62

Outline of Lecture

Outline of Lecture

• Term „deformation“

• Virtual works principle

• Deformation of bar – axial loading

• Deformation of bar – transversal loading

• Deformation of bar – torsional loading

• Deformation of indirect bar

• Deformation of curved bar

• Deformation of plane truss structure

• 3 / 62

Deformation

Term „deformation“

Deformation:

a) Global deformation of structure

b) Local component of deformation in some point (displacement, rotation)

• 4 / 62

Deformation

Why to calculate deformations?

1. Usability of structure

2. Solution of statically indeterminate structures

3. Verifying the correctness of the calculation by measurement

Calculation assumptions:

 Physical linearity (Hooke's law applies)

 Geometric linearity (small deformations theory)

Consequence:

 Equilibrium conditions are formulated on the deformed structure –

The First order Theory

 Apply the principle of superposition and the principle of

proportionality

Term „deformation“

• 5 / 62

Deformation

Nonlinear mechanics:

 2nd order theory – equilibrium conditions formulated on deformed

structure (small deformations)

 Physical nonlinearity (nonlinearly elastic or permanent deformations)

 Theory of big deformations

 Structures with unilateral links

 Cable structures

Term „deformation“

• 6 / 62

Work of external forces and moments

Virtual works principle

Work of point force and point moment

 cos ce PPLWork (external) of a force at point:

Work - scalar, units are Joules (J = N.m), kJ, MJ

.MLe Work of a moment at point:

Notice:

It is assumption that  () has

other cause than P (M).

The work is positive when there are same

directions of:

 vectors of force and displacement ,

 moment and rotation .

• 7 / 62

Work of continuous force and moment loading

Work of continuous loading

( ) ( ) b

a

e xxwxqL d ( ) ( ) b

a

xe xxxmL d

Assumption – magnitude of loading is constant during movement.

Work of external forces and moments:

Virtual works principle

• 8 / 62

Virtual work

a) Deformational virtual work

b) Force virtual work

1) Real loading state

2) Virtual loading state

2a) Deformational virtual state

2b) Force virtual state

ce

ce

wPL

wPL





Deformational virtual work described by Lagrange

to study equilibrium of structures

Virtual works principle

real loading state real deflection curve

virtual deflection curve force virtual

loading state

• 9 / 62

Work of internal forces

Coordinate system of the bar

Loaded bar in a space: N, My, Mz, Vz, Vy, T

Virtual works principle

• 10 / 62

Work of internal forces

Work of internal forces of bar

  

  

  l

x

l

y

l

z

l

zz

l

yy

l

i TvVwVMMuNL  dˆdˆdddd

Positive directions of internal forces

Work of internal forces:

Internal forces restrain deformations, they have opposite direction compared

to picture below, that is reason for negative sign in calculation of Li.

Virtual works principle

• 11 / 62

Virtual works principle

0 ie LL

Axiom:

Total virtual work on solved structure (i.e. sum of works of

external and internal forces) is equal to zero.

A) Deformational principle of virtual works (principle of virtual displacements)

B) Force principle of virtual works (principle of virtual forces)

Virtual internal forces

Real internal forces, causes deformations

x EA

N u dd 

x GA

V w

z

z dˆd *

x EI

M

y

y

y dd 

x GA

V v

y

y dˆd

* 

x EI

M

z

z z dd 

x GI

T

t

x dd 

TVVMMN yzzy ,,,,,

Virtual works principle

• 12 / 62

Deformational loading caused by temperature

Uniform thermal loading and decomposition of linearly changing thermal loading

across cross-section

Force principal of virtual works

   

   

 

 

  



  



l

tztyt

l

ty

yy

z

zz

z

zz

y

yy

e x b

t M

h

t MtNx

GI

TT

GA

VV

GA

VV

EI

MM

EI

MM

EA

NN L

0

21 0

0

** dd 

dxtdu

h

e tttt

t

z hdh





0

0 )(

 h

dx td

ttt

ty

hd





1

1



Virtual works principle

• 13 / 62

Betti's theorem (1872)

Enrico Betti

(1823 - 1892)

The work done by the 1st loading

state through the displacements

produced by the 2nd loading

state is equal to the work done

by the 2nd loading state the

displacements produced by the

2nd loading state.

 l

y

yy x

EI

MM PP

0

II,I,

2211 d

 l

y

yy x

EI

MM MP

0

I,II,

4433 d

44332211  MPPP 

Virtual works principle

1st loading state

2nd loading state

• 14 / 62

Maxwell’s theorem

Zvláštní případ Bettiho věty, kdy v každém z obou zatěžovacích

stavů působí na konstrukci jediná síla P nebo jediný moment M.

James Clerk

Maxwell

(1831 - 1879)

Displacement done by the first force in the place and direction of

second force is equal to displacement done by second force in the

place and direction of the first force.

IIIIII  PP  PPP  III   III

A special case of Betti's theorem. In each loading state acts only

one force P or moment M.

Virtual works principle

1st state

2nd state

• 15 / 62

Unit force method

Unit force method

  .1eL  

  



  



l

ty

yy

z

zz

z

zz

y

yy x

GI

TT

GA

VV

GA

VV

EI

MM

EI

MM

EA

NN

0

** d

   

   

 



l

tztyt x b

t M

h

t MtN

0

21 0 d

Force loading

Thermal loading

Virtual works principle

• 16 / 62

Deformation of bar – axial loading

Deformation of bar – axial loading

Deformation of bar exposed to axial loading

 l

e x A

NN

E u

0

d 1

Nt

l

te AtxNtu 0 0

0 d   

Force loading

Thermal loading

EA

A xNN

EA u N

l

e

  

0

d 1

Constant cross-section

Variable cross-section

Simpson’s rule ( ) ( ) 

3 24d 42310

0

d fffffxxf

l



• 17 / 62

Example 2.1

xN

R

R

ax

ax

.4,813

kN13

085,2.4,8





Deformation of bar – axial loading

Problem definition and solution of example 2.1

A = 64 mm2,

E = 2,1.108 kPa, t = 1,2.10 -5K-1

Calculate horizontal displacement uc for force and thermal loading state

Force loading state:

m000685,0 10.4,6.10.1,2

2,9

d

58

0





 l

N c

EA

A x

EA

NN u

• 18 / 62

Example 2.1

Deformation of bar – axial loading

Problem definition and solution of example 2

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