Introduction to statically indeterminate...

Department of Structural Mechanics

Faculty of Civil Engineering, VŠB-Technical University of Ostrava

Statics of Building Structures I., ERASMUS

Introduction to statically

indeterminate structures

2 / 58

Outline of Lecture

Outline of Lecture

• Statically indeterminate structures, degree of

indeterminacy

• Force method

• Simple statically indeterminate beam

• Simple statically indeterminate beam – axial loading

• Unilaterally fixed beam – transversal loading

• Bilaterally fixed beam – transversal loading

• Simply supported beam as an element of statically

indeterminate structure

• Simple statically indeterminate beam – torsional loading

• Calculation of deformation of the simple statically

indeterminate beam

3 / 58

• free mass point in the plane: nv=2,

coordinates [x, y], 2 different positions

• free mass point in the space:

nv=3, coordinates [x, y, z], 3 different

positions

• free solid bar in the plane:

nv=3, coordinates [x, y, ], 3 different

positions

• solid body in the space: nv=6,

coordinates [x, y, z, ], 6 different

positions

Possibility of free motion of physical objects

+x

+z

m[xm,zm]

x’

z’

Degree of freedom nv : ability to make one right-angled component of

displacement or rotation.

Statics of structures – basic knowledge

4 / 58

External constraints to movement

(a) (b) (c) (d) (e)

(f) (g)

(a) (b) (c) (d) (e)

Displacement constraint – prevents movement of constrained point in the

given direction


5 / 58

External constraints to rotation

Complex displacement and rotation constraints

Simple rotation constraints

(a) (b) (c)

(a) (b) (c)

Rotation constraint – prevents rotation of constrained bar in the given

direction

Full fix (clamp) in

space or plane, partial

fix in plane


6 / 58

Multiplicity of constraints

External constraints eliminates degrees of freedom of the object.

Name Multiplicity of

constraint

Symbol of constraint

(support), reactions

Pin, swing rod

Roller

Hinge

Sliding clamp

Fixed (clamped)

Examples of simple constraints and their reactions of solid bar in the plane

n–multiple constraint eliminates n degrees of freedom (n=1, 2, 3) of

object

a

Raz

aRaz

aRaz Rax

a Raz Rax May

a Raz May

1

2

2

3

1


7 / 58

Ensuring immobility of the bar

To sufficiently support some object it is needed v constraints to eliminate nv

degrees of freedom.

v = nv

Supporting of the object is kinematically determinate and

statically determinate, immobility of building is ensured,

structure can be used as building construction.

v < nvSupporting of the object is kinematically indeterminate and

statically overdeterminate, immobility of building is not

ensured, structure is inadmissible as the building construction.

(insufficient number of constrained degrees of freedom)

v > nvSupporting of the object is kinematically overdeterminate

and statically indeterminate, immobility of building is

ensured, structure can be used as building construction..

Constraints must be properly arranged to ensure immobility of structures –

it must not be Exceptional case of kinematically determinate or

overdeterminate structure.


8 / 58

Kinematically and statically determinate structure

Supporting of the structure is kinematically and

statically determinate

b

Rbz

a

Raz

Rax

P1 P2

Raz

RaxMay

P1 P2

a

v = nv

v = 3, nv = 3


9 / 58

Statically and kinematically determinate bars

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

(l)

nv = 6

nv = 3

nv = 3

nv = 2

Transversal loading

nv = 1

Axial loading

nv = 1

Torsional loading

nv = 3

nv = 3

nv = 3

nv = 3

nv = 3

nv = 6


10 / 58

Statically indeterminate structure

Supporting of the structure is kinematically

overdeterminate and statically indeterminate

b

Rbz

a

Raz

Rax

P1 P2

Raz

RaxMay

P1 P2

a

v = 4

nv = 3Rbx

Rbz

Rbx

Mby

b

v = 6

nv = 3

v > nv


11 / 58

Kinematically indeterminate structure

b

Rbz

a

Raz

P1 P2

Equilibrium of the object is ensured only for specific

loading

In building practice unusable structure

Supporting of the structure is kinematically

indeterminate and statically overdeterminate

v < nv


12 / 58

Exceptional cases of supporting

Constraints must be properly arranged to ensure immobility of structures

– it must not be exceptional case of supporting that is unusable in

construction practice.

b Rbxa

Raz

Rax

P1 P2

P1 P2

c

Rcz

a

Raz Rbz

b


13 / 58



(a)

(b)

(c)

(d)

(e)

(c) beam is not secured against rotation

(d) 3 constraints against movement intersect at one point

(e) 3 constraints against movement lying in one direct line


14 / 58

Equilibrium conditions of loaded bar

Number of equilibrium conditions depends on the type of solved task. It is

equal to degree of freedom of unsupported bar nv.

Supported bar has to be motionless and in equilibrium.

Number of reactions is equal to number of constrained degrees of freedom v.

v = nv Number of unknown reactions is equal to number of

equilibrium conditions, the bar is statically determinate

v < nv

v > nv

Number of unknown reactions is smaller than number of

equilibrium conditions, the bar is kinematically indeterminate

and cannot be used as building construction.

Number of unknown reactions is bigger than number of

equilibrium conditions, the bar is statically indeterminate, it can

be building construction. There must be supplemented

deformation conditions to statical equilibrium conditions.

If the determinant of system of equations is equal to zero, then it is exceptional case

of supporting of structure.


15 / 58

Equilibrium conditions of loaded bar system

For every separate bar can be written 3 equilibrium conditions

Number of reactions is equal to number of constrained degrees of freedom

of a structure.

Number of internal and external constraints: v = ve + vi


16 / 58

Kinematic and static determinacy of truss

Truss as a system of mass-points and

external and internal links

evps.2Condition of kinematical determinacy:

Practical approach – computational model is made from mass-points (at

joints) and internal links (bars), that constrain mutual displacement of both

connected joints.


bar (internal link) joint (mass-point)

external constraint

(support)

17 / 58

Kinematic and static determinacy

Raz

Raxa

Rbz

b

F3F2F1

N1 N5 N9

N3 N7 N11

N2 N6 N10

N4 N8

c d

e f g

s=7 number of joints (2 equilibrium conditions in each of them)

p=11 number of bars (1 unknown axial force in each of them)

a1=1

a2=1number of constraints with 1 and 2 unknown reactions

14.2.2 21 aaps


18 / 58


Raz

Rax

a

Rbz

b

N1

N3

N2

c d

s=4

p=5

a1=1

a2=1

8.2 s 8.2 21 aap

N4

N5

Statically and kinematically determinate truss

Kinematically indeterminate truss

F2F1

s.221 .2 aap


19 / 58


s=4

p=6

a1=0

a2=2

8.2 s 10.2 21 aap 2x statically indeterminate truss

Raz

Rax

a

Rbz

b

N1

N3

N2

c d

N4

N5

N6

This is not joint

(bars are

overlapping

here)

F2F1

Rbx


20 / 58

Calculation of the degree of static indeterminacy

Plane frame structures and beams

1. Open bar systems:

ns = v – 3 – pk = a1 + 2.a2 + 3.a3 – 3 – pk

v number of external constraints (reactions)

ai number of external constraints with 1, 2 or 3 reactions

pk number of internal hinges recalculated to simple

hinges

2. Closed bar systems:

ns = 3.u + v – 3 – pk

u number of closed areas

21 / 58

Force method

Force method

Force method: • is used for solution of statically indeterminate structures, ns ≥ 1,

• is a basic method for solution of statically indeterminace bar

structures

• is a direct method

Force method utilizes• equilibrium conditions

• deformational conditions

• principal of superposition and principal of proportionality

22 / 58

Force method

Routine of solution of structures by the Force method:

1. Determination of degree of statical indeterminacy ns

2. Creating of Basic statically determinate structure by

removing of ns constraints (external or internal)

3. Removed constraints are replaced by statically

indeterminate forces or moments (statically indeterminate

reactions or interactions)

4. Formulating of the system of ns deformational conditions

in the form of linear equations

5. Solving of the system of linear equations. The solution are

values of statically indeterminate reactions or interactions.

6. Calculation of remaining reactions and internal forces

Force method

23 / 58

Simple statically indeterminate beam


Assumptions:

a) direct beam of constant or variable cross-section

b) axis of the beam is identical with x-axis

c) beam is supported in 2 points,

d) every of external constraints is parallel to one of

coordinate axis,

e) every of external rotational constraints works in the

plane that is perpendicular to some of the coordinate

axis

f) a beam prut can be spatially loaded

24 / 58



Spatial problem of simple direct beam

The degree of statical indeterminacy ns = v – nv is number

of redundant constraints, i.e. number of constraints that

has to be removed to make the beam statically determinate

25 / 58



Every problem of a simple statically indeterminate beam

in a space (3D) can be divided into 4 simplifier

problems:

1. Axial problem…nv=1

2. Transversal problem in xz plane…nv=2

3. Transversal problem in xy plane…nv=2

4. Torsional problem…nv=1

26 / 58

Simple statically indeterminate beam – axial loading

ll

bx

bx

vs

dxAE

dxAE

NN

RX

X

RX

nvn

0

1

0

11

11

101

10111

1

11

0

112

Silové zatížení

lll

dxA

N

Edx

AE

NNdx

AE

NN

0

0

0

10

0

010

1


Force method in axial problem

Silové

zatížení

Silové zatížení

Oteplení

ltdxNt t

l

t 0

0

1010

27 / 58


Displacement of supports

Displacement in x-axis direction

Deformational condition: 110111 dX

In this case (X1 parallel to x-axis):

ūa and Ra1 have oposite direction

ūb and X1 have same direction

b

aaaa

ud

uuuRR

1

110 )1()()(

,au bu

110

11

1

111

XRRR

Ruu

X

uuX

axaxax

bxab

ba

Force method in axial problem

28 / 58

Example 3.1

kNEA5

10681,9

)(9,2)9,1(18,4

)(9,1

)(9,13

7,5

10681,9

7,5

10681,9

2

122,1

2

11)2,48,4(

10681,9

1

10681,9

3

)(1

)(8,4

0

110

11

101

5510

3

0

05

0

010

5

0

111

1

0

10111

kNXRRR

kNR

kNXR

dxNdxAE

NN

AE

ldx

AE

NN

R

kNR

X

axaxax

bx

bx

l

l

a

a

Force loading


Problem definition and solution

of the example 3.1

Deformational condition

29 / 58

Example 3.1



of the example 3.1

Thermal loading

The beam is loaded by change of temperature t0=15oC. The temperature is constant

along length of beam.

)(258,174 )(258,174

)(258,174)258,174(10

)(258,1743

10681,9104,5

10681,9

3

104,5315102,1

0

15 ,1021 ,10681,9

110

54

11

10

511

45

00

0

10

10111

0

0

55

kNRkNR

kNXRRR

kNR

EA

l

ltdxtN

X

C Δt,αkNEA

bxax

axaxax

bx

tt

l

-

t

Deformational condition:

30 / 58

Example 3.1, supports shifting

Ūa= 5 mm = 0,005 m (to the right)

Ūb= 8 mm = 0,008 m (to the right)

Displacement Ūa and reaction Ra1 are in opposite direction

Displacement Ūb and unit force X1=1 are in same direction


)(1,968 )(1,968

)(1,968

0

)(1,96810681,93

005,0008,0

008,0

005,0)1()()(

10681,9

3

5

11

1

1

110

511

kNRkNR

kNRR

RR

kNuu

XR

ud

uuuRR

EA

l

bxax

bxax

bxax

abbx

b

aaaax

110111 dX


of the example 3.1

31 / 58

Transversally loaded statically indeterminate beams

110111

10111

:supports of shiftingby loading -condition nalDeformatio

0

:loading thermaland loading force -condition nalDeformatio

123 :acyindetermin statical of Degree

dX

X

nvn vs

Simple statically indeterminate beam – transversal loading

One-end fixed statically indeterminate beam

32 / 58

Bilaterally fixed beam – transversal loading

Force method - bilaterally fixed beam with

transversal loading

224vs nvn

baab MXMX

XX

XX

21

20222112

10212111

0

0

Deformational conditions for

loading by lifting of supports:

220222112

110212111

dXX

dXX

Deformational conditions for force

loading and thermal loading:

33 / 58

Example 3.8

The beam of steel cross-section I200 is loaded by:

forces according to pict. (a)

linear change of temperature t1=15o K – pict. (e)

supports shifting – pict. (h)

Deformational conditions for force and thermal loading:

0

0

20222112

10212111

XX

XX

Deformational conditions for loading by supports shifting:

220222112

110212111

dXX

dXX


of the example 3.8

34 / 42

Example 3.8 – solution, force loading

Force loading

Basic statically determinate structure: simply supported beam

Statically indeterminate values: X1 = Mab X2 =- Mba

„0“ loading state:

Reactions Raz0, Rbz0:

Bending moment Mo:

Can be also written:

lx xxx

qMxxFM

lx x

qMxRM

MMMM

xxqxxF

xqxRxM

xx

qxRxM

kNR

kNR

F

F

qFF

qazRa

qFqRa

F

Faz

az

bz

az

for 2

)( )(

0for 2

where

M

4,8x1,6for

2

)()(

2)(

6,10for 2

)(

6,278,4/)8,06,166,1142,32,310(

288,4/)6,12,3102,31446,16(

2

220

2

11000

2001000

2

2

2

100

2

100

0

0

35 / 42


beam supportedsimply of angles naldeformatio basic , , ,

8,0

6

8,0

6

8,4

6

section -crossconstant for EI

1,6

3

6,1

3

8,4

3

:tscoefficien naldeformatio ofn Calculatio

1 state loading 2nd

1 state loading1st

ba ab

0

12

21

0

21

12

0

22

22

0

11

11

2

2

1

1

EIEI

ldx

IE

MM

EIEIEI

ldx

IE

MM

EI

ldx

IE

MM

EIEIEI

ldx

IE

MM

xlM

X

xM

X

l

ba

l

ab

baab

l

ba

l

ab

36 / 42


Calculation of deformational coefficients 10, 20

14044,58

30222,60

: isit case In this

2.2 Table of use with 3)

rule, sin'Vereshchag of use with 2)

y,analytical 1) :realized becan n Integratio

)(1

)(1

20

10

2000

0

0

0

20

20

1000

0

0

0

10

10

21

21

EI

EI

dxMMMMMEI

dxEI

MM

dxMMMMMEI

dxEI

MM

qFq

l

R

l

qFq

l

R

l

a

a

37 / 42


Solution of linear equations is:

kNR

XRXRRR

kNR

XRXRRR

kNmX

kNmX

b

bbbb

a

aaaa

037,278,4

3244,23

8,4

0266,266,27

563,288,4

3244,23

8,4

0266,2628

:Reactions

32444,23

02667,26

22110

22110

2

1


of the example 3.8

38 / 42

Example 3.8 – thermal loading

0)8,4

1

8,4

1(0

alsoor ,0 and . because ,0Reaction

(g). pict. see -moment bending of Diagram

4494 supplied For

0446,44,2

10216

8,06,1

10216

102162,0

15102,1

0)(

:one into reduced becan equationslinear twoofSolution

thereforel,symmetrica is problem The

15tsection -cross theacross re temperatuof changeLinear

210

2

55

1211

10

5

8,4

0

5

0

1

110

101211

21

1

XXRXRRR

VkonstMRR

kNmEIEI

kNmEI

EIEI

X

dxl

xldx

h

Mt

X

XXX

C

azazazaz

bzaz

l

t

o


of the example 3.8

39 / 42

Example 3.8 – loading by supports shifting

(j) and (i) pictures see -Solution

144,68,4

)491,18(

8,4

)001,11(0

18,491kNm-MX -11,001kNM :is equations ofSolution

10625,41025,6004,04494

6,1

4494

8,0

1025,64494

8,0

4494

6,1

:onsubstitutiAfter

1025,68,4

006,0003,0)

8,48,4()()(

1025,68,4

006,0003,0)

8,48,4()()(

direction) same have and ( 004,0 ,0 :is system edeterminat statically basic selectedFor

:conditions nalDeformatio

22110

ab2ab1

44

20221

4

10121

4

2220

4

1110

b121

220222121

110212111

kNXRXRRR

X

dXX

dXX

wwwRwRR

wwwRwRR

Xdd

dXX

dXX

abababab

ba

bbaa

ba

bbaa

Problem definition and solution of the example 3.8

40 / 58

Table 3.2

Unilaterally fixed beam – transversal loading

Fixed-end moment reactions of

the beam with constant cross-

section

41 / 58

Simply supported beam as an element of statically indeterminate

structure

Simply supported beam as an element of statically indeterminate structure

Decomposition of simply supported beam into separate loading states

42 / 58

Simply supported beam as an element of statically indeterminate

structure – diagrams of internal forces

Simply supported beam as an element of statically indeterminate structure

Diagrams of bending moments in loading state “0” and in “moment’s” loading state

Moment’s loading state:0. loading state (q): 0. loading state (P):

43 / 58

Variable cross-section of the beam


44 / 58

Variable cross-section - example 3.5

kNmnxq

nx

baRM

mq

baR

nx

kNl

abMlq

abR

kNl

abMlq

baR

abMl

ql

baR

kNmlqab

M

71,162

2

545,114545.163,21

2

2

,,max

545,114

25,26,

:max

M ofn Calculatio

37,485

85,66

2

514,

2, similarly

63,215

85,66

2

514,

2,

0,

2

2, :reacions ofn Calculatio

85,6625141910,02

,

1910,0 obtained becan ion interpolatlinear by ,1855,0 ,1921,0 tableFrom

125,00,6

0,3 4,0

5

2

:3.3 Tableby Solution

15,012,0

3

3


45 / 58

Table 3.3

Coefficients

for moment

reaction

(uniformly

distributed

loading)


46 / 58

Table 3.4


Coefficients

for moment

reaction

(nodal force

loading)

47 / 58

Table 3.6


Coefficients

for moment

reaction

(uniformly

distributed

loading)

48 / 58

Table 3.7


Coefficients

for moment

reaction

(nodal force

loading)

49 / 58

Force method and torsional loading

Simple statically indeterminate beam – torsional loading

110111

10111

vs

dδXδ

:shifting supportsby loadingfor b)

0δXδ

:loading forcefor a)

condition nalDeformatio

112nνn

:acyindetermin of Degree


50 / 58

Torsional problem – example 3.11

2

4433

9013

1075,936,024,0196,0

: tab.)see ( is stiffness Torsional

kNmIG

mhbI

t

t

h/b 1,0 1,1 1,2 1,3 1,4 1,5

0,1406 0,1540 0,1661 0,1771 0,1869 0,1958

h/b 1,6 1,7 1,8 1,9 2 3

0,2037 0,2109 0,2174 0,2233 0,2287 0,2633

Problem definition:

Reinforced-concrete beam (G=9,24.106kPa) of constant rectangular cross-

section, width b=0,24m, height h=0,36m. It is loaded by:

a) torsional loading according pict. (a)

b) supports shifting rad. 002,0 rad, 001,0 ba



of the example 3.11

51 / 58

Torsional problem – example 3.11, force loading

(d) pict. see - T moments torsionalof diagram Resulting

25,675,7114

75,74

31

312

5,13

2

5,2)914(

1

4

moments) loading tocompareddirection oposite has (X 1XMoment

0

:condition nalDeformatio

110

1

11

10

1

4

0

0

0

10

10

0

11

11

11

10111

kNmXTTT

XTkNmIG

IGX

IGIGdxT

IGdx

IG

TT

IGIG

ldx

IG

TT

X

aaa

b

t

t

ttt

l

t

l

ttt



of the example 3.11

52 / 58

Torsional problem – example 3.11, supports shifting

kNm 76,60

kNm 76,610438,4

003,0

10438,4

)001,0(002,0

rad 002,0

rad 001,0))001,0(1()(

rad/kNm10438,49013

44

44

11

1011

1

110

4

11

110111

baba

b

aa

t

TTTT

dTX

d

T

GI

dX

Supports shifting:


rad. 002,0 rad, 001,0 ba



of the example 3.11

53 / 58

Calculation of deformation of simple statically indeterminate beam

Force loading state

Reduction theorem:When calculation deformation of statically indeterminate structure of degree of indeterminacy ns than the unite force loading state can be made on:

original statically indeterminate beam

statically indeterminate beam with lower degree of indeterminacy nsj < ns, (removed less than ns links),

statically determinate beam (removed ns links).

Reduction theorem can be used to calculate the deformation of any statically indeterminate structure exposed to force loading.

Calculation of deformation of the simple statically indeterminate beam

54 / 58Calculation of deformation of the simple statically indeterminate beam

Calculation of deformation – force loading state

Examples of virtual load case options to calculate the displacement of

statically indeterminate beam.

55 / 58

Example 3.12 and its solution

Calculation of deformation – force loading state, example 3.12

2.2 Table using c)

rule sin'Vereshchag using b)

lyanalytical a)

:solved becan above formula The

1296

51

(c). pictureon as choosed state forceunit Virtual

3.2, tableusing calculated weremoments Bending

s..point in beam theofnt displaceme Calculate

3

0IE

PldxMM

IEdx

IE

MMw

ll

s


56 / 58

Calculation of deformation – force loading state, example 3.12

EI

PlPl

IEw

PlPl

IEw

llPlllPl

IEw

MAMAIE

w

dxMMIE

dxIE

MMw

s

s

s

TMTMs

ll

s

1296

5

12183

)149(1

)12183

7

818(

1

)23

2

2

1

254

7

4218(

1

)(1

1

33

33

2211

0

Calculation using Vereshchagin’s rule



of the example 3.12

57 / 58

Calculation of deformation – thermal loading

tpr w ww

Resulting deformation is given by superposition of:

a) elastic deformation wpr

b) thermal deformation wt on basic statically determinate

structure


58 / 58

Example 3.13 and its solution

Calculation of deformation – thermal loading, example 3.13

mw

mdxh

Mtw

mw

EIw

ww

kNmEI

c

l

t

t

pr

pr

tpr

333

35

0

1

3

c

2

10931,010326,310395,2

10326,32

232,15

2,0

18102,1

10395,2

4494/)2

2,2135,2232,1562.4

2

8,2232,1(

1

w

4494 :Solution

The beam is loaded by change of temperature.

Calculate displacement in point c.


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