Analysis of Statically Indeterminate Structures by the …fast10.vsb.cz/koubova/DSM_truss.pdf · 2017. 10. 27. · Analysis of Statically Indeterminate Structures by the Displacement

Analysis of Statically

Indeterminate

Structures by the

Displacement Method

Displacement method

1. Slope-Deflection Method

� In this method it is assumed that all deformations

are due to bending only.

� Deformations due to axial forces are neglected.

2. Direct Stiffness Method

� Deformations due to axial forces Deformations due to axial forces Deformations due to axial forces Deformations due to axial forces are notnotnotnot neglectedneglectedneglectedneglected.

Displacement method

1. Slope-Deflection Method

� The continuous beam is kinematically indeterminate to

second degree.

� Two unknown joint rotation ϕϕϕϕb, ϕϕϕϕc

2. Direct Stiffness Method

� The continuous beam is kinematically indeterminate to

fourth degree.

� Two unknown joint rotation ϕϕϕϕb, ϕϕϕϕc. and two translations ub, uc

a

q

b c

F1 F2

d

Direct Stiffness Method:

Plane Truss

Truss Analysis

� Plane trusses are made up of short thin members

interconnected at hinges to form triangulated patterns

� A hinge connection can onlyonlyonlyonly transmittransmittransmittransmit forcesforcesforcesforces from one

member to another member but notnotnotnot thethethethe momentmomentmomentmoment

� For analysis purpose, the truss is loadedloadedloadedloaded atatatat thethethethe jointsjointsjointsjoints

� Hence, a truss member is subjected to onlyonlyonlyonly axialaxialaxialaxial forcesforcesforcesforces

and the forces remain constant along the length of the

member

� The forces in the member at its two ends must be of the

same magnitude but act in the opposite directions for

equilibrium

Truss Analysis

� Consider a truss member having crosscrosscrosscross sectionalsectionalsectionalsectional areaareaareaarea A,

Young’sYoung’sYoung’sYoung’s modulusmodulusmodulusmodulus of material E, lengthlengthlengthlength of the member l

� Let the member be subjected to axial tensile force F

� Under the action of constant axial force, applied at each

end, the member gets elongationelongationelongationelongation u

� The force-displacement relation for the truss member

may be written:

� k is the stiffnessstiffnessstiffnessstiffness ofofofof thethethethe trusstrusstrusstruss membermembermembermember and is defined as

the force required for unit deformation of the structure

EA

Flu =F F

u

ukul

EAF ⋅=⋅=

Direct Stiffness Method: Plane Truss

Example 1Example 1Example 1Example 1

F2 = 20 kN

F1 = 10 kN

a b

c

EA = const.

3 3

4

1. Degrees of freedom

� Truss is kinematically

indeterminate to

second degree.

� Two translations are

unknown – horizontal uc

and vertical wc


Example 1Example 1Example 1Example 1 F2 = 20 kN

F1 = 10 kN

a b

c

3 3

4

2. Express normal forces

due to elongation:

bcbc

cbbc

acac

caac

abab

baab

ul

EANN

ul

EANN

ul

EANN

⋅==

⋅==

=⋅== 0

( )

5

3cos

5

4sin

m543 22

==

=+==

αα

bcac ll

α α



3. Equilibrium equations (write one equilibrium equation

for each unknown translation)

0sinsin

0coscos

0sinsin

0coscos

0

0

2

1

2

1

=⋅⋅+⋅⋅+

=⋅⋅+⋅⋅−

=⋅+⋅+

=⋅+⋅−

=

=

∑

∑

αα

αα

αα

αα

bcbc

acac

bcbc

acac

cbca

cbca

z

x

ul

EAu

l

EAF

ul

EAu

l

EAF

NNF

NNF

F

F

F2

F1

c

Nca

αα

Ncb



3. Equilibrium equations

205

4

55

4

5

105

3

55

3

5

sinsin

coscos

2

1

−=⋅⋅+⋅⋅

−=⋅⋅+⋅⋅−

−=⋅⋅+⋅⋅

−=⋅⋅+⋅⋅−

bcac

bcac

bcbc

acac

bcbc

acac

uEA

uEA

uEA

uEA

Ful

EAu

l

EA

Ful

EAu

l

EA

αα

αα

EAu

EAu

EAuu

EAuu

bc

ac

bcac

bcac

⋅−=

⋅−=

⋅−=+

⋅−=+−

6

625

6

125

4

500

3

250



kN833,206

625

5

kN167,46

125

5

−=

⋅−⋅=⋅==

−=

⋅−⋅=⋅==

EA

EAu

l

EANN

EA

EAu

l

EANN

bcbc

cbbc

acac

caac

4. Normal forces(After evaluating elongations, substitute it to evaluate normal

forces.)



kN333,35

4167,4

sin:0

kN5,25

3167,4

cos:0

=⋅+=

⋅−==

=⋅+=

⋅−==

∑

∑

az

caazza

ax

caaxxa

R

NRF

R

NRF

α

α

a

Nac = Nca

αNab = 0

Rax

Raz

b

Nbc = Ncb

αNba = 0

Rbx

Rbz

5. Reactions

kN667,165

4833,20

sin:0

kN5,125

3833,20

cos:0

=⋅+=

⋅−==

=⋅+=

⋅−==

∑

∑

bz

cbbzzb

bx

cbbxxb

R

NRF

R

NRF

α

α

Direct Stiffness Method:

Plane Truss

Matrix NotationMatrix NotationMatrix NotationMatrix Notation

Direct Stiffness Method, Matrix NotationMatrix NotationMatrix NotationMatrix Notation

Local and Global CoLocal and Global CoLocal and Global CoLocal and Global Co----ordinate Systemordinate Systemordinate Systemordinate System

� All members are oriented at

different directions

� It is required to transform

member displacements and

forces from the locallocallocallocal cocococo----

ordinateordinateordinateordinate systemsystemsystemsystem x*z* to

globalglobalglobalglobal cocococo----ordinateordinateordinateordinate systemsystemsystemsystem xz

� So that a global load-

displacement relation may

be written for the complete

truss

F2

F1 x

z

x*

z*


Transformation matrix Tab

⋅

=

⇒+=+=

ba

ba

ab

ab

ba

ab

bababa

ababab

Z

X

Z

X

X

X

ZXX

ZXX

γγγγ

γγγγ

sincos00

00sincos

sincos

sincos*

*

*

*

⋅=

ba

ba

ab

ab

ab

ba

ab

Z

X

Z

X

X

XT

*

*

=

γγγγ

sincos00

00sincosabT

T1abab TT =−

ab

ab

ab

ab

l

xx

l

zz

−=

−=

γ

γ

cos

sin

x

z

Xab

Zab

Xba

Zba

γ γ γ γ … … … … angle of angle of angle of angle of

transformation transformation transformation transformation

a

b

orthogonal matrix

−−

⋅=

−

−=

⋅=

⋅

−

−=

=

−

11

11*

*

**

*

*

*

*

ab

abab

ababab

b

aab

b

a

abab

abab

ba

ab

ba

ab

l

EA

l

EA

l

EAl

EA

l

EA

u

u

u

u

l

EA

l

EAl

EA

l

EA

X

X

N

N

k

k


Member local stiffness matrix kab*


Member global stiffness matrix kab*

−−−−

−−−−

=

⋅

−

−⋅

=⋅⋅=

⋅=

⋅⋅⋅=

⋅⋅=

⋅=

2

22

22

22

22

*T

*T

*

**T

*

*T

sscssc

sccscc

sscssc

sccscc

sincos00

00sincos

sin0

cos0

0sin

0cos

abab

abab

abababababab

b

b

a

a

ab

b

b

a

a

ababab

b

aabab

ba

abab

ba

ba

ab

ab

l

EA

l

EA

l

EAl

EA

l

EA

w

u

w

u

w

u

w

u

u

u

X

X

Z

X

Z

X

k

TkTk

kTkTkTT

γγγγ

γγ

γγ


Example 1Example 1Example 1Example 1, , , , Matrix NotationMatrix NotationMatrix NotationMatrix Notation

F2 = 20 kN

F1 = 10 kN

a b

c

EA = const.

3 3

4

Member local stiffness matrix:

−−

⋅=

−−

⋅=

−−

⋅=

−−

⋅=

−−

⋅=

−−

⋅=

11

11

511

11

11

11

511

11

11

11

611

11

*

*

*

EA

l

EA

EA

l

EA

EA

l

EA

cbcb

acac

abab

k

k

k



Transformation matrix:

6,05

36cos

8,05

04sin

6,05

03cos

8,05

40sin

16

06cos

06

44sin

=−=−=

=−=−=

=−=−=

−=−=−=

=−=−=

=−=−=

cb

cbcb

cb

cbcb

ac

acac

ac

acac

ab

abab

ab

abab

l

xx

l

zz

l

xx

l

zz

l

xx

l

zz

γ

γ

γ

γ

γ

γ

a b

c

EA = const.

3 3

4

x

z

[0 0] [3 0]

[0 4][6 4]



Transformation matrix:

=

=

−−

=

=

=

=

8,06,000

008,06,0

sincos00

00sincos

8,06,000

008,06,0

sincos00

00sincos

0100

0001

sincos00

00sincos

cbcb

cbcbcb

acac

acacac

abab

ababab

γγγγ

γγγγ

γγγγ

T

T

T




⋅

−−

⋅⋅

=⋅⋅=

−−

⋅

−−

⋅⋅

−

−=⋅⋅=

⋅

−−

⋅⋅

=⋅⋅=

8,06,000

008,06,0

11

11

5

8,00

6,00

08,0

06,0

8,06,000

008,06,0

11

11

5

8,00

6,00

08,0

06,0

0100

0001

11

11

6

00

10

00

01

*T

*T

*T

EA

EA

EA

bcbcbcbc

acacacac

abababab

TkTk

TkTk

TkTk




−−−−

−−−−

⋅=

−−−−−−

−−

⋅=

−

−

⋅=

64,048,064,048,0

48,036,048,036,0

64,048,064,048,0

48,036,048,036,0

5

64,048,064,048,0

48,036,048,036,0

64,048,064,048,0

48,036,048,036,0

5

0000

0101

0000

0101

6

EA

EA

EA

bc

ac

ab

k

k

k



Member vector of unknown forces:

⋅

−−−−

−−−−

⋅=

⋅=

⋅

−−−−−−

−−

⋅=

⋅=

=

⋅

−

−

⋅=

⋅=

0

0

64,048,064,048,0

48,036,048,036,0

64,048,064,048,0

48,036,048,036,0

5

0

0

64,048,064,048,0

48,036,048,036,0

64,048,064,048,0

48,036,048,036,0

5

0

0

0

0

0

0

0

0

0000

0101

0000

0101

6

c

c

b

b

c

c

cb

bc

bc

cb

cb

c

c

c

c

a

a

ac

ca

ca

ac

ac

b

b

a

a

ab

ba

ba

ab

ab

w

u

EA

w

u

w

u

Z

X

Z

X

w

u

EA

w

u

w

u

Z

X

Z

X

EA

w

u

w

u

Z

X

Z

X

k

k

k



The load displacement equation for the truss:

S … the vector of joint loads acting on the truss

r … the vector of joint displacements

K … the global stiffness matrixglobal stiffness matrixglobal stiffness matrixglobal stiffness matrix

∑

∑

=+=

=+=

2

1

:0

:0

FZZF

FXXF

cbcazc

cbcaxcF2 = 20 kN

F1 = 10 kN

c

SrK =⋅

=

⋅

++−+−+

⋅

=

⋅

⋅+

⋅

−−

⋅

20

10

64,064,048,048,0

48,048,036,036,0

5

20

10

64,048,0

48,036,0

564,048,0

48,036,0

5

c

c

c

c

c

c

w

uEA

w

uEA

w

uEA



The load displacement equation for the truss:

=

⋅=

⋅

−

=⋅=

=

⋅=

++−+−+

⋅

−

=

c

c

w

u

EAEA

EA

EAEA

125,78

444,691

20

10

1

144,00

0144,0

20

10

144,00

0144,0

64,064,048,048,0

48,048,036,036,0

5

SKr

S

K

1



Member vector of joint displacements

=

=

=

=

=

=

=

⋅=

0

0

125,78

444,69

0

0

125,78

444,69

0

0

0

0

125,78

444,691

EA

EA

w

u

w

u

EA

EA

w

u

w

u

w

u

w

u

w

u

EA

b

b

c

c

cb

c

c

a

a

ac

b

b

a

a

ab

c

c

rr

r

r



Member vector of knownknownknownknown globalglobalglobalglobal forces:

−−

=

⋅

−−−−

−−−−

⋅=⋅=

−−

=

⋅

−−−−−−

−−

⋅=⋅=

=

⋅

−

−

⋅=⋅=

67,16

5,12

67,16

5,12

0

0

125,78

444,69

64,048,064,048,0

48,036,048,036,0

64,048,064,048,0

48,036,048,036,0

5

33,3

5,2

33,3

5,2

0

0

125,78

444,69

64,048,064,048,0

48,036,048,036,0

64,048,064,048,0

48,036,048,036,0

5

0

0

0

0

0

0

0

0

0000

0101

0000

0101

6

EA

EAEA

Z

X

Z

X

EA

EAEA

Z

X

Z

X

EA

Z

X

Z

X

cbcb

bc

bc

cb

cb

acac

ca

ca

ac

ac

abab

ba

ba

ab

ab

rk

rk

rk



Member vector of knownknownknownknown locallocallocallocal forces:

−=

−−

⋅

=

⋅=

−=

−−

⋅

−−

=

⋅=

=

⋅

=

⋅=

833,20

833,20

67,16

5,12

67,16

5,12

8,06,000

008,06,0

167,4

167,4

33,3

5,2

33,3

5,2

8,06,000

008,06,0

0

0

0

0

0

0

0100

0001

*

*

*

*

*

*

bc

bc

cb

cb

cb

bc

cb

ca

ca

ac

ac

ac

bc

ac

ba

ba

ab

ab

ab

ba

ab

Z

X

Z

X

X

X

Z

X

Z

X

X

X

Z

X

Z

X

X

X

T

T

T



Check: Check: Check: Check:

� forces equilibrium equation

a b

c

Xac

ZacXac

*

Xca*

Xca

ZcaXcb*

Xbc*

Zcb

Xcb

Zbc

Xbc

F2

F1

2067,1633,3

105,125,2

:0

:0

2

1

=+=+−

=+=

=+=

∑∑

FZZF

FXXF

cbcazc

cbcaxc



Normal forces:Normal forces:Normal forces:Normal forces:

−

=

−=

−

=

−=

−

=

=

bc

cb

bc

cb

ca

ac

bc

ac

ba

ab

ba

ab

N

N

X

X

N

N

X

X

N

N

X

X

833,20

833,20

167,4

167,4

0

0

*

*

*

*

*

*

a b

c

Xac

Zac-Nac = Xac

*

Nca = Xca*

Xca

Zca-Ncb = Xcb*

Nbc = Xbc*

Zcb

Xcb

Xbc

F2

F1

Zbc

kN833,20

kN167,4

−==

−==

cbbc

caac

NN

NN



RRRReactionseactionseactionseactions::::

a b

c

Rax = Xac

-Raz = ZacXac

*

Xca*

( )( )( )( )↑=−=

←=−=

↑=−=

→==

kN67,16

kN5,12

kN33,3

kN5,2

bcbz

bcbx

acaz

acax

ZR

XR

ZR

XR

Xca

ZcaXcb*

Xbc*

Zcb

Xcb

-Rbx = Xbc

F2

F1

-Rbz = Zbc


Example Example Example Example 2222

Degrees of freedom:

� Truss is kinematically

indeterminate to 7th

degree.

� Seven translations are

unknown – horizontal

ub, uc, ud, ue and

vertical wc, wd, we.

eF1 = 4kN

F2 = 8kN

a b

c d

1

4

5

6

2 2

3

323

7

E = 20 GPaA1,4,5,6 = 0,02 m2

A2,3,7= 0,01 m2



Code number:

� Non-zero code number is assigned code number is assigned code number is assigned code number is assigned to each unknown

translation.

7

6

5

4

3

2

1

=

b

d

d

c

c

e

e

u

w

u

w

u

w

u

r

e

a b

c d

1

4

5

6

23

7

(1 2)

(3 4) (5 6)

(0 0) (7 0)



Member parameters:

( ) m61,332 227,3,2 =+=l

Member E [kPa] A l cos sin

1 (ab) 20000000 0,02 4 1 0

2 (db) 20000000 0,01 3,606 0,555 0,832

3 (ad) 20000000 0,01 3,606 0,555 -0,83

4 (ca) 20000000 0,02 3 0 1

5 (cd) 20000000 0,02 2 1 0

6 (ec) 20000000 0,02 3 0 1

7 (ed) 20000000 0,01 3,606 0,555 0,832

ab

abab

ab

abab

l

xx

l

zz

−=

−=

γ

γ

cos

sin

eF1

F2

a b

c d

1

4

5

6

2 2

3

323

7

[0 0]

[0 3]

[0 6]

x

z

[2 3]

[4 6]




−

−=

l

EA

l

EAl

EA

l

EA*k

k1* 100000,0 -100000,0

-100000,0 100000,0

k2* 55470,0 -55470,0

-55470,0 55470,0

k3* 55470,0 -55470,0

-55470,0 55470,0

k4* 133333,3 -133333,3

-133333,3 133333,3

k5* 200000,0 -200000,0

-200000,0 200000,0

k6* 133333,3 -133333,3

-133333,3 133333,3

k7* 55470,0 -55470,0

-55470,0 55470,0



Member transformation matrix:

=

γγγγ

sincos00

00sincosT

T1 1,00 0,00 0,00 0,00

0,00 0,00 1,00 0,00

T2 0,55 0,83 0,00 0,00

0,00 0,00 0,55 0,83

T3 0,55 -0,83 0,00 0,00

0,00 0,00 0,55 -0,83

T4 0,00 1,00 0,00 0,00

0,00 0,00 0,00 1,00

T5 1,00 0,00 0,00 0,00

0,00 0,00 1,00 0,00

T6 0,00 1,00 0,00 0,00

0,00 0,00 0,00 1,00

T7 0,55 0,83 0,00 0,00

0,00 0,00 0,55 0,83



Member global stiffness matrix: TkTk ⋅⋅= *T

0 0 7 0 code number 5 6 7 0 code number 0 0 5 6 code number

k1 100000 0 -100000 0 0 k2 17067,7 25601,55 -17067,7 -25601,5 5 k3 17067,7 -25601,5 -17067,7 25601,55 0

0 0 0 0 0 25601,55 38402,32 -25601,5 -38402,3 6 -25601,5 38402,32 25601,55 -38402,3 0

-100000 0 100000 0 7 -17067,7 -25601,5 17067,7 25601,55 7 -17067,7 25601,55 17067,7 -25601,5 5

0 0 0 0 0 -25601,5 -38402,3 25601,55 38402,32 0 25601,55 -38402,3 -25601,5 38402,32 6

3 4 0 0 code number 3 4 5 6 code number 1 2 3 4 code number

k4 0 0 0 0 3 k5 200000 0 -200000 0 3 k6 0 0 0 0 1

0 133333,3 0 -133333 4 0 0 0 0 4 0 133333,3 0 -133333 2

0 0 0 0 0 -200000 0 200000 0 5 0 0 0 0 3

0 -133333 0 133333,3 0 0 0 0 0 6 0 -133333 0 133333,3 4

1 2 5 6 code number

k7 17067,7 25601,55 -17067,7 -25601,5 1

25601,55 38402,32 -25601,5 -38402,3 2

-17067,7 -25601,5 17067,7 25601,55 5

-25601,5 -38402,3 25601,55 38402,32 6



Global stiffness matrix (partial calculation):

� “Localization” “Localization” “Localization” “Localization” according to the code numberMember 1 1 2 3 4 5 6 7 Member 5 1 2 3 4 5 6 7

1 12 23 3 200000 0 -200000 04 4 0 0 0 05 5 -200000 0 200000 06 6 0 0 0 07 100000 7

Member 2 1 2 3 4 5 6 7 Member 6 1 2 3 4 5 6 71 1 0 0 0 02 2 0 133333,3 0 -1333333 3 0 0 0 04 4 0 -133333 0 133333,35 17067,7 25601,55 -17067,7 56 25601,55 38402,32 -25601,5 67 -17067,7 -25601,5 17067,7 7

Member 3 1 2 3 4 5 6 7 Member 7 1 2 3 4 5 6 71 1 17067,7 25601,55 -17067,7 -25601,52 2 25601,55 38402,32 -25601,5 -38402,33 34 45 17067,7 -25601,5 5 -17067,7 -25601,5 17067,7 25601,556 -25601,5 38402,32 6 -25601,5 -38402,3 25601,55 38402,327 7

Member 4 1 2 3 4 5 6 7123 0 04 0 133333,3567



Global stiffness matrix for truss (summation of partial calculations)

K 1 2 3 4 5 6 7

1 17067,7 25601,55 0 0 -17067,7 -25601,5 0

2 25601,55 171735,7 0 -133333 -25601,5 -38402,3 0

3 0 0 200000 0 -200000 0 0

4 0 -133333 0 266666,7 0 0 0

5 -17067,7 -25601,5 -200000 0 251203,1 25601,55 -17067,7

6 -25601,5 -38402,3 0 0 25601,55 115207 -25601,5

7 0 0 0 0 -17067,7 -25601,5 117067,7



Vector of joint loads acting on the truss:

eF1 = 4kN

F2 = 8kN

a b

c d

1

4

5

6

2 2

3

323

7

=

=

0

0

0

0

8

0

4

xb

zd

xd

zc

xc

ze

xe

F

F

F

F

F

F

F

S


Example Example Example Example 2222The load displacement equation for the truss:

SKrSrK 1 ⋅=⇒=⋅ −

7

6

5

4

3

2

1

000080,0

000105,0

000392,0

000045,0

000432,0

000090,0

000918,0

−

−

=

=

b

d

d

c

c

e

e

u

w

u

w

u

w

u

r

code n. Member1 code n. Member2 code n. Member3 code n. Member4 code n. Member5 code n. Member6 code n. Member7

0 0,000000 5 0,000392 0 0,000000 3 0,000432 3 0,000432 1 0,000918 1 0,000918

0 0,000000 6 0,000105 0 0,000000 4 -0,000045 4 -0,000045 2 -0,000090 2 -0,000090

7 0,000080 7 0,000080 5 0,000392 0 0,000000 5 0,000392 3 0,000432 5 0,000392

0 0,000000 0 0,000000 6 0,000105 0 0,000000 6 0,000105 4 -0,000045 6 0,000105

Member vector of joint displacements:

� Creating according to the code number



Member vector of known global forces:

Member vector of known local forces:

Member1 Member2 Member3 Member4 Member5 Member6 Member7

0 -8 5 8 0 -4 3 0 3 8 1 0 1 4

0 0 6 12 0 6 4 -6 4 0 2 -6 2 6

7 8 7 -8 5 4 0 0 5 -8 3 0 5 -4

0 0 0 -12 6 -6 0 6 6 0 4 6 6 -6

abab

ba

ba

ab

ab

Z

X

Z

X

rk ⋅=


-8,00 14,42 -7,21 -6,00 8,00 -6,00 7,21

8,00 -14,42 7,21 6,00 -8,00 6,00 -7,21

⋅=

ba

ba

ab

ab

ab

ba

ab

Z

X

Z

X

X

XT

*

*



Normal forces:Member N [kN]

1 (ab) 8,00

2 (db) -14,42

3 (ad) 7,21

4 (ca) 6,00

5 (cd) -8,00

6 (ec) 6,00

7 (ed) -7,21

−

=

ba

ab

ba

ab

N

N

X

X*

*


-8,00 14,42 -7,21 -6,00 8,00 -6,00 7,21

8,00 -14,42 7,21 6,00 -8,00 6,00 -7,21

Member vector of known local forces:



Reactions:

( ) ( )

( ) kN12120

kN12660

kN12048

−=−+=

+=

+=++=

++=

−=+−+−=

++=

bz

bdbabz

az

acadabaz

ax

acadabax

R

ZZR

R

ZZZR

R

XXXR

eF1 = 4kN

F2 = 8kN

a b

c d

1

4

5

6

23

7

Rax

Raz Rbz

Member1 Member2 Member3 Member4

Xab -8 Xdb 8 Xad -4 Xca 0

Zab 0 Zdb 12 Zad 6 Zca -6

Xba 8 Xbd -8 Xda 4 Xac 0

Zba 0 Zbd -12 Zda -6 Zac 6

Member vector of known global forces

Documents

Analysis of Statically Indeterminate Structures by the …fast10.vsb.cz/koubova/DSM_truss.pdf · 2017. 10. 27. · Analysis of Statically Indeterminate Structures by the Displacement