statics of building structures i., erasmusfast10.vsb.cz/koubova/sobsi_theme3_frames.pdfvierendeel...

Department of Structural Mechanics

Faculty of Civil Engineering, VŠB-Technical University of Ostrava

Statics of Building Structures I., ERASMUS

Frame structure

• Basic properties of plane frame structure

• Simple open frame structure

• Simple closed frame structure

2 / 52Basic properties of plane frame structure

Examples of simple open plane frame

Types of plane frames

Frames: a) right-angled

b) oblique

c) branched

d) open (a), (b), (c)



Frames: a) right-angled

b) oblique

c) branched

d) closed (a), (b), (c)

Examples of simple closed plane frame


Branched frame


5 / 52

Examples of right-angled and oblique coupled frames

Types of plane frames, coupled frames

Basic properties of plane frame structure

Coupled frames – originates by combining several simple

open frames

6 / 52

Vierendeel truss and Storey frame


Basic properties of plane frame structure

Vierendeel truss – originates by combining several closed

frames side by side

Storey frame - originates by combining several closed frames

above each other

7 / 52Simple open frame structure

The first step of the Force method

Force method, simple open frame

8 / 52

Different ways to create a basic statically determinate structure within the second

step of the Force method


Simple open frame structure

9 / 52

Replacement of removed links by reactions or interactions in the third step of

the Force method



10 / 52

Equation in pictures illustrating decomposition into loading states


0

0

0

:loading thermaland forcefor

conditions nalDeformatio

30333232131

20323222121

10313212111

XXX

XXX

XXX


11 / 52


j

m

j

l

j

j

itj

m

j

l

jit

j

m

j

l

j

i

j

m

j

l

j

i

ikki

j

m

j

l

j

ki

j

m

j

l

j

ki

iki

dxh

tMdxtN

dxAE

NNdx

IE

MM

mdxAE

NNdx

IE

MM

jj

jj

jj

1 0

,1

1 0

,0i,0

1 0

0

1 0

0

i,0

,,

1 01 0

ki,

s0,

n

1k

k,

s

:loading thermal todue tscoefficien naldeformatio ofn Calculatio

:loading force todue tscoefficien naldeformatio ofn Calculatio

frame theof bars ofnumber is

:tscoefficien naldeformatio ofn Calculatio

)1,.....n i(for X

:formin structure ateindetermin statically timesnfor

written becan equations) (canonical conditions nalDeformatio

s

12 / 52

Force method, simple open frame, support shifting

Deformational conditions for support shifting:

s0,

1

,

330333232131

220323222121

110313212111

n1,...,ifor ii

n

k

kki dX

dXXX

dXXX

dXXX

s

13 / 52

Calculation of deformational coefficients due to support shifting


vuMuHu

lwMwRw

M

udwdd

ww

aaaaaab

aaaaaab

aaa

bb

a

ba

b

b

)(

)(

)(

)(clockwise),(clockwise

),(u ),(u ),( ),( :shifts of directions and shiftingSupport

33

*

30

22

*

20

1

*

10

321

b

ba


14 / 52


330333232131

220323222121

110313212111

dXXX

dXXX

dXXX

vuuXXX

lwwXXX

XXX

vulw

udwdd

aab

aab

ab

aaaa

abbb

333232131

323222121

313212111

3020

10321

is

,

, , , ,

15 / 52

Two constraints in the axis of the same bar

Warning


Bar “c-d” is supported against movement in the direction of

the axis of the bar.

It is necessary to take into account the influence of normal

forces on the deflection of the bar “c-d”.

Otherwise, the system of canonical equations singular.

16 / 52

Example 5.1

I1=0,002m2, I2=I3=0,004m2

6,03,5

2,1cos ,8,0

5,3

8,2sin

033,59)33333,1(1,2

8,2

5,38,21,2

0

22

1,

arctgarctg

mll ca


17 / 52

Loading states and bending moment diagrams for loading states of Example 5.1

Example 5.1, solution

)(1H ),(7,5

8,2 ),(

7,5

8,2

0H ),(7,5

1 ),(

7,5

1

0H ,0 ),(30

a222

a111

a000

kNRkNR

kNRkNR

RkNR

ba

ba

ba


18 / 52

Example 5.1, solution continuation

EE

EE

EEE

EEE

EEE

4,64989

3

5,376842.163

.002,0

1

6,41585)

3

36842,063158,0(

2

5,363

.002,0

1

5,2762

3

6,376842,1

.004,0

1

3

5,376842,1

.002,0

1

4,15026,3

004,03

63158,076842,1)

3

36842,063158,0(

2

5,376842,1

.002,0

1

1,1304

004,03

6,363158,0))36842,0

3

263158,0(

2

5,3036842

2

63158,15,363158,0(

.002,0

1

20

10

22

22

2112

2

11

0

0

20222121

10212111

XX

XX

Deformational conditions:


19 / 52

Example 5.1, solution of linear equations

kNX

kNX

XX

XX

EEEEE

906,165,1345370

2274471

4,15024,15025,27621,1304

4,1502)6,41585()4,64989(1,1304

814,125,1345370

17240145

4,15024,15025,27621,1304

4,649894,1502)4,64989(1,1304

04,649895,27624,1502

06,415854,15021,1304

4,64989 ,

6,41585 ,

5,2762 ,

4,1502 ,

1,1304

2

1

21

21

201022211211

0

0

20222121

10212111

XX

XX

Deformational conditions:


20 / 52

Example 5.1, completion, reactions and internal forces diagrams

)(557,16

)(381,10)557,16(7,5

8,2)814,12(

7,5

10

clockwise)(counter 814,12

)(557,16)557,16(100

)(381,40)557,16(7,5

8,2)814,12(

7,5

130

2

22110

1

22110

22110

kNXH

kNXRXRRR

kNmXM

kNXHXHHH

kNXRXRRR

b

bbbb

a

aaaa

aaaa


21 / 52

Removal of internal links and its replacement by interactions

Simple closed frame


22 / 52

First three steps of the Force method

Simple closed frame

.

3acy indetermin statical of Degree

constIE

ns


23 / 52

Loading states and corresponding bending

moments diagrams for Example 5.2

Example 5.2, loading states

)(8

)(33320

)(6669

czech)in stav" "0. (i.e. "0" state loading

in theonly reactions Nonzero

0

0

0

kNRR

kN,RR

kN,RR

axax

bzbz

azaz


Notice to bending moment diagram (see pic.):

“Bottom side” of horizontal bars is down.

“Bottom side” of vertical bars is considered at

the right side of bar.

24 / 52

Example 5.2, canonical equations

IEIE

IEIE

IEIE

IEIE

XXX

XXX

XXX

732,78))7,2(6,3)7,2(22

3

7,2)7,2(2

3

7,2)7,2((

1

088,101)6,34,56,3

3

6,36,3

3

6,3)6,3((

1

0 0

40,32))6,3(14,5

2

6,316,3)

2

6,3(16,3(

1

18)116,3)1()1(6,3114,5)1()1(4,5(

1

0

0

0

22

33

22

22

32233113

2112

11

30333232131

20323222121

10313212111

Calculation of deformational coefficients:

25 / 52

Example 5.2, canonical equations

kNVVXkNNNXkNmMMX

XXX

XXX

XX,X

IEIE

IEIE

IEIE

edecedecedec 667,2 ,602,8 ,209,2

:

952,20932,7870 0

012,7980 088,1014,32

95,238 0 432 18

952,209))7,2(

2

6,3)8,28(

6

9,547,2)7,2()

3

8,289,548,28(

2

7,27,2(

1

12,798))6,3(6,3

2

)8,28(6,37,2

2

9,546,37,2

2

9,548,28(

1

95,238)16,3

2

)8,28()1(7,2

2

9,54)1(7,2

2

9,548,28(

1

321

321

321

321

20

20

10

equations canonical ofSolution

:equations canonicalin on Substituti

:tscoefficien naldeformatio ofn Calculatio

26 / 52

Example 5.2, completion

Internal forces can be determined:

a) From the equilibrium conditions with

knowledge of reactions and statically

indeterminate forces

b) By superposition of loading states,

taking into account real value of

statically indeterminate forces (see

bellow)

Ad b):

xxxxx

xxxxx

xxxxx

NXNXNXN

VXVXVXVV

MXMXMXMM

3322110

3322110

3322110

N


27 / 52


X1=-2,209kNm, X2=-8,602kNm, X3=-2,667kN

26,141kNmM

9,409kNmM 9,409kNmM

4,991kNmM 4,991kNmM

21,5573kNmM 21,5573kNmM

7,159kNmM (column) 7,159kNmM

m ax

dbdc

cdca

babd

abac

)0(667,2)6,3(602,8)1()209,2(9,54

)7,2(667,2)0(602,8)1()209,2(0

)7,2(667,2)0(602,8)1()209,2(0

)7,2(667,2)6,3(602,8)1()209,2(0

)7,2(667,2)6,3(602,81)209,2(8,28

max

3322110

M

M

M

M

M

MXMXMXMM

dc

ca

bd

ac

xxxxx


28 / 52

Example 5.2, bending moments calculation

X1=-2,209kNm, X2=-8,602kNm, X3=-2,667kN

26,141kNmM

9,409kNmM 9,409kNmM

4,991kNmM 4,991kNmM

21,5573kNmM 21,5573kNmM

7,159kNmM (column) 7,159kNmM

m ax

dbdc

cdca

babd

abac

)0(667,2)6,3(602,8)1()209,2(9,54

)7,2(667,2)0(602,8)1()209,2(0

)7,2(667,2)0(602,8)1()209,2(0

)7,2(667,2)6,3(602,8)1()209,2(0

)7,2(667,2)6,3(602,81)209,2(8,28

max

3322110

M

M

M

M

M

MXMXMXMM

dc

ca

bd

ac

xxxxx


29 / 52

Example 5.2, another way to calculate bending moments

X1=-2,209kNm, X2=-8,602kNm, X3=-2,667kN

kNmM

VMM

kNmMM

kNmM

NMM

kNM

NVMM

kNmMM

kNmMM

ba

ababba

acab

ac

eccaac

ac

ececceac

cdca

cecd

559,21

7,2304,5333,12159,77,2304,5

159,7

159,7

6,3602,86,38991,46,36,38

159,76,3602,87,2667,26,38209,2

6,37,26,38

991,4

991,47,2667,2209,2

:better

It is necessary to know some components

of internal forces in this shortened

calculation (e.g. shear force Vab)


30 / 52

Resulting reactions, interactions and diagrams of internal forces for Example 5.2.



31 / 52

Steel frame structure of industrial hall

Span 20,5 m

Examples of frame structures

32 / 52

Hall for manufacturing components for nuclear power plants,

Vitkovice

• Ground130 x 320 m

• Cranes with capacity 80 and 200 t

• Undermined area


33 / 52

Steel frame structure of coupled hall, Vítkovice

• Span 30 a 24 m

• Cranes with capacity 80 a 50 t

• Undermined area


34 / 52

Sport hall Slavia, Prague


35 / 52

Administration Building, Glasgow, UK


Space steel frame with bracing

36 / 52



Space steel frame with bracing

37 / 52



Space steel frame with bracing - detail

38 / 52

San Sebastian, Auditorium, Spain


Space frame

39 / 52

San Sebastian, Auditorium, Spain


40 / 52

Congress Centre, Brno Exhibition Centre

Visible space frame supporting structure


41 / 52

University Children's Hospital, Brno

Supporting space frame structure with overhanging ends


42 / 52

Elementary School, Brumov – Bylnice

Frame structure with bracing


43 / 52

Aula, VŠB-TU Ostrava


Space frame of reinforced concrete

44 / 52

Aula, VŠB-TU Ostrava


Space frame of reinforced concrete - detail

45 / 52

Radio Free Europe, Prague

Vierendeel truss of

1968:

• Ground 59x83 m

• 6 pillars


46 / 52


Vierendeel truss of

1968:

• Ground 59x83 m

• 6 pillars


47 / 52


Vierendeel truss of

1968:

• Ground 59x83 m

• 6 pillars


48 / 52

Road bridge, Karviná – Darkov Spa

RC arched bridge of 1925:

• Vierendeel truss

• Unique cross-bracing

• Height 6,25 m

• Deck length 55,8 m

• Width 6,25 m


Photo: Ing. Renata Zdařilová

49 / 52


Reinforced Concrete

arched bridge of

1925:



50 / 52




Reinforced Concrete

arched bridge of

1925:

51 / 52



52 / 52




statics of building structures i., erasmusfast10.vsb.cz/koubova/sobsi_theme3_frames.pdfvierendeel...

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