Deformations of statically determinate bar...

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 statereal deflection curve

virtual deflection curveforce 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

xEA

Nu dd

xGA

Vw

z

z dˆd*

xEI

M

y

y

y dd

xGA

Vv

y

ydˆd

*

xEI

M

z

zz dd

xGI

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 xb

tM

h

tMtNx

GI

TT

GA

VV

GA

VV

EI

MM

EI

MM

EA

NNL

0

210

0

**dd

dxtdu

h

etttt

t

zhdh

0

0 )(

h

dxtd

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

yyx

EI

MMPP

0

II,I,

2211 d

l

y

yyx

EI

MMMP

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

yyx

GI

TT

GA

VV

GA

VV

EI

MM

EI

MM

EA

NN

0

**d

l

tztyt xb

tM

h

tMtN

0

210 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 xA

NN

Eu

0

d1

Nt

l

te AtxNtu 0

0

0 d

Force loading

Thermal loading

EA

AxNN

EAu N

l

e

0

d1

Constant cross-section

Variable cross-section

Simpson’s rule( ) ( )

324d 42310

0

dfffffxxf

l

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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:

m000685,010.4,6.10.1,2

2,9

d

58

0

l

Nc

EA

Ax

EA

NNu

18 / 62

Example 2.1

Deformation of bar – axial loading

Problem definition and solution of example 2.1

mm48,0m00048,02).20.(10.2,1

dd

5

0

0

0

0

0

c

Nt

l

t

l

tc

u

AtxNtxtNu

Displacement caused by thermal

loading:

19 / 62

Example 2.2

Deformation of bar – axial loading

Problem definition and solution of example 2.2

Concrete

r = 2400 kg.m-3

E = 2.107 kPa

Calculate vertical

displacement wb of

top of column for

loading by self-weight

20 / 62

Example 2.2

Deformation of bar – axial loading

Problem definition and solution of example 2.2

E

Zz

z

zz

E

zA

N

Ez

EA

NNw

i

ni

i i

ii

b

1

2

4

0

4

0

.2,08,0

.4,2.2,191

d1

d

2.4,2.2,19

2)..8,42,192,19()(

zz

zzzN

zz

zA .2,08,0)4

.8,08,0.(1)(

33kNm24Nm240002400.10

z

zAzn

.8,42,19

24)..2,08,0()(

21 / 62

Example 2.2

Solution using: 1) Simpson’s rule

1

4

0

kNm895,1543

)7240.2)571,56600,21.(40(d

z

A

N

Deformation of bar – axial loading

i z A N N/A

m m2 kN kNm-2

0 0 0,8 0 0,.0000

1 1 1 -21,6 -21,.6000

2 2 1,2 -48 -40,0000

3 3 1,4 -79,2 -56,5714

4 4 1,6 -115,2 -72,0000

3)2)(4(d)( 4231

4

0

0

dfffffxxf 1

4

4d

mm007745,0m10.745,710.2

895,154 6

7

bw

22 / 62

Example 2.2

Solution using:

m10.749,710.2

9756,154

4,010

4

10

6

7

E

Zw

n

lzz

n

b

i

Deformation of bar – axial loading

2) Rectangular method

(numerical integration)

i zi Ni /Ai

m kNm-2

1 0,2 1,874286

2 0,6 5,384348

3 1,0 8,64

4 1,4 11,69778

5 1,8 14,59862

6 2,2 17,3729

7 2,6 20,04364

8 3,0 22,62857

9 3,4 25,14162

10 3,8 27,59385

S Ni /Ai 154,9756

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Axially loaded bar – column

Structure with axially loaded bar

Graded cross-section of column on tall building, Chicago, USA

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Deformation of bar – transversal loading

Deformation of direct beam – transversal loading

Types of transversally loaded direct beams

ll

xA

VV

Gx

I

MM

E0

*

0

d1

d1

l

t xM

t0

1 dh

Force loading

Thermal loading

ll

xVVGA

xMMEI

0

*

0

d1

d1

Constant

cross-section

25 / 62

Vereshchagin’s rule

TM

l

MAxMM .d0

Aid for computation of the integral

Deformation of direct beam – transversal loading

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Vereshchagin’s rule

Parabolic parts of moment diagrams for use within Vereshchagin’s rule

Deformation of direct beam – transversal loading

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Example 2.3

Problem definition and solution of example 2.3

Reinforced concrete cantilever

E = 2,2.107 kPa

Calculate vertical

deflection = wa. Use

Vereshchagin’s rule.

Neglect the work of

shear forces.

Deformation of direct beam – transversal loading

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Example 2.3

Problem definition and solution of example 2.33

3

1

0

33

3

2

2

1

22

3

1

1

0

11

3

321

2

0

23

37

3

kNm667,21)667,1.(13

d

kNm15)5,1.(10

d

kNm5,2)75,0.(333,3

d

m005407,010.24416,7

667,21155,2

)(1

d

kNm1024416,7

12

28,0.18,0.10.2,2

12

MAxMMS

MAxMMS

MAxMMS

SSSEI

xEI

MMw

bhEEI

M

M

M

a

Deformation of direct beam – transversal loading

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Example 2.4

Real and virtual shear forces in example 2.4

Reinforced concrete cantilever

G = 9,24.106 kPa

The same problem as in Example 2.3 but the

work of shear forces is taken into account

00

´

5

5

´

2,2

1,1

56*

2*

21*

2

0

*

´

7,1100.407,5

093,0100.

mm093,0m10.276,910.8808,3

2610

kNm26)1.(1.26

kNm10)1.(2

1.20

kN10.8808,3042,0.10.24,9

m042,02,1

28,0.18,0

)(1

w

w

w

VAS

VAS

GA

bhA

SSGA

dxGA

VVw

c

c

V

V

c

Deformation of direct beam – transversal loading

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Table 2.2

Equations for calculation of

integrall

xMM0

d

Deformation of direct beam – transversal loading

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Example 2.5

Problem definition and solution of example 2.5

Calculate vertical

displacement wc

and rotation a

Wood

E = 107 kPa

Deformation of direct beam – transversal loading

32 / 62

Table 2.3

Local deformations of a cantilever beam and simply supported beam

Deformation of direct beam – transversal loading

33 / 62

Example 2.6

Problem definition and solution of example 2.6

M

t

l

t Ah

txM

h

t 1

0

1 d

Steel t = 1,2.10-5 K-1

Thermal loading – the change of temperature is linear along height of the cross-section

Calculate deflections wc a ws.

h = 0,24 m

Deformation of direct beam – transversal loading

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Example 2.6

Problem definition and solution of example 2.6

cM

tttc A

h

txM

h

tx

h

Mtw 1

9

0

9

0

11 dd

92

2.9

cMA

mm2,7m0072,024,0

)9.(16.10.2,15

cw

sM

ts A

h

tw 1

125,6

2

75,1.7

sMA mm9,4m0049,0

24,0

125,6.16.10.2,15

sw

Deformation of direct beam – transversal loading

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Example 2.6

Problem definition and solution of example 2.7

Shape, loading – see Example 2.3

Variable cross-section

Reinforced concrete cantilever

E = 2,2.107 kPa

Deformation of direct beam – transversal loading

36 / 62

Research energetic center, VŠB-TU Ostrava

Example of the structure with variable cross-section

Cantilever beam:

• Steel welded and rolled I- profile

• Trapeze metal plate

• Concrete floor

37 / 62

Research energetic center, VŠB-TU Ostrava

Example of the structure with variable cross-section

Cantilever beam:

• Steel welded and rolled I- profile

• Trapeze metal plate

• Concrete floor

38 / 62

Deformation of bar – torsional loading

t

T

l

t

l

t

cGI

AxTT

GIx

GI

TT

00

d1

d

Torsional rotation

Force virtual state

Deformation of direct beam – torsional loading

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Example 2.8

Determine torsional rotation of

the right end b. Use unit force

method.

Steel - G = 8,1.107 kPa

o

2

00

277

4544

4

1

4

2

20,2rad0384,08466,60

336,2

kNm336,26,0.72,0.2

11).52,172,2.(

2

1

d1

d

kNm8466,6010.5119,7.10.1,8

mm10.5119,7)2430.(2

)(2

b

T

t

T

l

t

l

t

t

pt

A

GI

AxTT

GIx

GI

TT

GI

rrII

Deformation of direct beam – torsional loading

Problem definition and solution of example 2.7

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Deformation of polyline beam – plane loading

Deformation of polyline beam – plane loading

m

j

l

j

j

l

j

j

l

j

j

jjj

xA

VV

Gx

I

MM

Ex

A

NN

E1 0

*

00

d1

d1

d1

3 local components of deformation: u, v a

m

j

l

j

j

j

xI

MM

E 1 0

d1

At most of statically determinate cases the work of shear and normal forces

is neglected

22

ccc uw At the point cc

c

w

utan

Thermal

loading

m

j

l

j

j

j

l

jjt

jj

xh

MtxNt

1 0

,1

0

,0 dd

Constant

cross-section

m

j

l

j

j

j

xMMIE 1 0

d11

41 / 62

Example 2.9

Problem definition and solution of example 2.9

Calculate

ud , wd , , d

Steel

I1 = 16.10-5 m4

I2 = 3,8.10-5 m4

I3 = 9,2.10-5 m4

E = 2,1.108 kPa

Deformation of polyline beam – plane loading

42 / 62

Steel frame structure of industrial hall

Examples of structures of polyline beams

Span 20,5 m

43 / 62

Industrial hall, Vítkovice

• Floor plan dimensions 130 x 320 m

• Cranes capacity 80 a 200 t

• Undermined region

Examples of structures of polyline beams

44 / 62

Multipurpose hall, Frýdek - Místek

Examples of structures of polyline beams

45 / 62

Multipurpose hall, Frýdek - Místek

Rámová ocelová konstrukce

Examples of structures of polyline beams

46 / 62

Platform of football stadium, Ostrava Bazaly

• Undermined region

Examples of structures of polyline beams

47 / 62

Platform of football stadium, Ostrava Bazaly

Detail of moment hinge

Examples of structures of polyline beams

48 / 62

Deformation of curved beam – plane loading

Deformation of curved beam – plane loading

Shape and supporting of a plane curved beam

Span l, deflection f, relative deflection Fl

fΦ

49 / 62

Deformation of curved beam – plane loading

Deflection and relative deflection on different curved beams.

l

fΦ

Deformation of curved beam – plane loading

Span l, deflection f, relative deflection F

50 / 62

Deformation of curved beam – plane loading

Force loading

L jLL

sGA

VVs

EI

MMs

EA

NNddd

*

Thermal loading

L

t

L

t sh

MtsNt dd 10

cos

dd

xs Solution After modification:

Force loading b

a

b

a

b

a

x

x

x

x

x

x

xA

VV

Gx

I

MM

Ex

A

NN

Ed

cos

1d

cos

1d

cos

1*

Thermal loading

b

a

b

a

x

x

t

x

x

t xh

Mtx

Nt d

cosd

cos10

Use of Unit force method

Deformation of curved beam – plane loading

51 / 62

Deformation of curved beam – plane loading

Calculation of deformation

Numerical integration

Simpson’s rule

Rectangular method

3))...(2)...(4(d)( 2421310

0

dffffffffxxf nnn

l

n

i

i

i

iin

i

i

i

ii

n

i

i

ii

iin

i

i

ii

ii

x

x

x

x

sI

MM

Es

A

NN

E

xI

MM

Ex

A

NN

Ex

I

MM

Ex

A

NN

E

b

a

b

a

11

11

11

cos

1

cos

1d

cos

1d

cos

1

Deformation of curved beam – plane loading

52 / 62

Example 2.10

Problem definition and solution of example 2.10

( ) 2.xkxz

22

b

b

a

a

x

z

x

zk

Parabolic central line

xkxkx

z..2.

d

dtg

2

2tg1

1cos

2tg1

tgsin

Calculate ub

EI = 6,72.104 kNm2

Deformation of curved beam – plane loading

53 / 62

Example 2.10

Problem definition and

solution of example 2.10

i x

[m]

tg

ψ

cos ψ M [kNm] [m] M/ cos ψ

[kNm2]

0 -

5,00

-

0,8

0,7808

7

0,0000 0,000 0,000

1 -

3,75

-

0,6

0,8574

9

28,4375 0,875 29,018

2 -

,2,5

0

-

0,4

0,9284

8

47,5000 1,500 76,739

3 -

1,25

-

0,2

0,9805

8

57,1875 1,875 109,350

4 0,00 0,0 1,0000

0

57,5000 2,000 115,000

5 1,25 0,2 0,9805

8

43,1250 1,875 82,461

6 2,50 0,4 0,9284

8

28,7500 1,500 46,447

7 3,75 0,6 0,8574

9

14,3750 1,875 14,668

8 5,00 0,8 0,7808

7

0,0000 0,000 0,000

Deformation of curved beam – plane loading

54 / 62

Curved beam

Gateway Arch, span of the steel arch from the year 1966 is 192,5 m, Saint Louis, Missouri.

Examples of planary loaded curved structures

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Gateway Arch, rozpětí a vzepětí ocelového oblouku z roku 1966 192,5 m, Saint Louis, Missouri.

Curved beam

Examples of planary loaded curved structures

56 / 62

Curved beam

Rovinně zakřivený vazník, Výzkumné energetické centrum VŠB-TU Ostrava

Examples of planary loaded curved structures

57 / 62

Deformation of plane truss structure

Deformation of plane truss

p

j j

jjjp

j

l

j

j

jjp

j

l

j

j

jj

A

lNN

Ex

A

NN

Ex

EA

NN jj

11 01 0

.1d

1d

Thermal loading

Virtual work of Normal forces only

p

j

jjt

p

j

l

jjt

p

j

l

jjt ltNxtNxtN

jj

1

,0

1 0

,0

1 0

,0 dd

58 / 62

Example 2.11

Problem definition and solution of example 2.11

Calculate wc

A1 = 24.10-4 m4

A2 = 12.10-4 m4

A3 = 18.10-4 m4

A4 = 18.10-4 m4

A5 = 12.10-4 m4

A6 = 12.10-4 m4

A7 = 18.10-4 m4

l2 = l3 = l6 = 2,236 m

Calculation in table

Deformation of plane truss

59 / 62

Example 2.11 Calculation in table

mm62,5m1062,5101,2

10192,11801 3

8

37

1

j j

jjj

cA

lNN

Ew

j Aj [m2] lj [m] Nj [kN] Ñj [1] (NjÑjlj/Aj).10-3 [kN/m]

1 0,0024 2,000 -90,000 -1,000 75,000

2 0,0012 2,236 134,164 2,236 559,017

3 0,0018 2,236 -67,082 0,000 0,000

4 0,0018 2,000 -60,000 -2,000 133,333

5 0,0012 1,000 0,000 0,000 0,000

6 0,0012 2,236 67,082 2,236 279,508

7 0,0018 2,000 -60,000 -2,000 133,333

1180,192

Deformation of plane truss

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Railway bridge, Polanka’s conjunction

from 1964

Examples of plane truss structures

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Railway bridge, Polanka’s conjunction

Examples of plane truss structures

62 / 62

Road bridge, Ostrava - Hrabová

Ukázky kloubových příhradových konstrukcí

Truss bridge over Ostravice river