Statically indeterminate examples - axial loaded members ...fast10.vsb.cz/lausova/indeterm_all.pdf · Statically determined and indetermined examples 2 / 28 Statically indeterminate

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Department of Structural MechanicsFaculty of Civil Engineering, VSB - Technical University Ostrava

Elasticity and Plasticity

Statically in determinate examples - axial loadedmembers , rod in torsion ,members in bending

2 / 28Statically determined and indetermined examples

Statically indetermin ate structures

Condition of solution:elastic (linear) behaviour of strain-stress diagram of material

Statically indetermined problems:

number of unknown variables

> number of equilibrium equations

= +number of

deformation conditions

Solution:

number of unknown variables

number of equilibrium equations

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Axially loaded members

1. Fixed supported column on both end

3. Nehomogenized bar (steel pipe filled in by concret).

2. Rods

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Example 1: Fixed supported column on both ends

1l

l

2l

F

aR

a

b

bR

Unknown variables in example:

( ) ( )21 , NRNR ba =−=

Equilibrium equation:0=−+ FRR ba:0=zR

Deformation equation:

0.. 22

22

11

1121 =+=∆+∆

AE

lN

AE

lNll

:0=∆l

Statically determined and indetermined examples

Condition of solution:elastic (linear) behaviour of strain-stress diagram of material

021 =−+− FNN

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Determine normal stress in both parts, cross sectio ns I140 and I180, F=650kN.

0)(

02

2

1

1

2

22

1

11 =−+≡=+EA

lFR

EA

lR

EA

lN

EA

lN bb

FN1

N2

Rb

-Ra

-

+

Rb

Ra

l1=1,5m

l2=2,5mI 180

I 140

2) Equilibrium equations (just axial task ):

0l =∆From the diagram ohf normal forces :

bRN =1 FRN b −=2

0lll 21 =∆+∆=∆⇒

∑ = 0,verticaliF

1) 1x statically indetermined in the axial task

3) Deformation condition:

By substituting into the deformation condition:

a)-( F -b2 RRN ===bR

2112

12

AlAlAl

F+

b1 RN =

Example 1

0=−+ FRR ba

We can determine the unknows just from the one equation – deformation condition.

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MPaA

Nx 03,186

1

11 ==σ

kNN 42,3112 =

kNRN b 58,3381 ==

FN1

N2

Rb

-Ra

-

+

Rb

Ra

l1=1,5m

l2=2,5mI 180

I 140

Normal stress in the bar:

MPaA

Nx 62,111

2

22 −==σ

Example 1

Determine normal stress in both parts, cross sectio ns I140 and I180, F=650kN.

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The bar is loaded by forces see the picture.A1 = 3cm2, A2 = 10cm2 , E1 = E2, F1 = 20kN, F2 = 45kN. • Guess the direction of the reactions and diagram of normal forces• Divide the bar on parts, where there will be different value of stresses and calculate them.

1

2F2

F1

0,6m

0,8m

0,4m

x

Example 2

Results.: N1=1,875kNN2=-18,125kN N3=26,85kN

Ra=26,85kN ↓ Rb=1,875kN ↑

σ1= 6,25MPa σ2= -18,125MPa σ3= 26,875MPa

a

b

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l=6m, E=2,1.105MPa,

NRR0RR baba ⇒=⇒=−

N0lT

EANl

l T =⋅∆α+=∆⇒

EATN T ⋅∆α−=⇒ kN15,255−=

[ ]151021 −−⋅=α C, oT

Ra Rb

ΔT

Determine normal stress in the bar U100, which is under the temperature change ∆T=90°C.

0=∆l

We can determine the unknows just from the one equation – deformation condition.MPa8,206

AN

x −==σ

4) Normal stress in the bar:

Example 2

2) Equilibrium equations (just axial task ):

∑ = 0,xiF



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Example 3

The bar of the lenght l = 1 m is between two stiff walls, the hole is 0,2 mm. What value of normal stress is in the rod, if the temperature change is +50°C?

1) Deformation equation:

m102,0 3−⋅=∆l

αT = 17 · 10-6 °C-1, E = 1,1 · 105 MPa

3T 102,0lT

EANl −⋅=⋅∆α+

0,2 mm

l=

1000

mm

2) Result:

MPa5,71x −=σ

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Example 4

g = 120 kNm-1

l=

12 m

a = 3 m b = 4 m

L = 7 m

a

b c

1 2

Determine stress in the rods, if both are from I 140.

Conditions of solution: elastic behaviour of the rods, ideally stiff slab

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∑ = 0Fiz

∑ = 0Fix

∑ = 0Mia

l=

12 m

a = 3 m b = 4 m

L = 7 m

a

b

Rb Rc

Raz

Rax

1 2

g

Example 4


Conditions of solution: elastic behaviour of the rods, stiff behaviour of the beam.

2) Equilibrium equations :

1) 1x statically indetermined


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(Equilibrium equations ):

∑ = 0Fiz∑ = 0Fix ∑ = 0M ia

l=

12 m

a b

a

b c

1 2

1l∆

2l∆

bal

al 21

+∆=∆

(2 unknowns forces N1, N2, choose an equation from Equilibrium equations, which includes just N1 a N2)

gN1 N2

( )ba1

EAlN

a1

EAlN 21

+=⇒


05,37q7N3N 21 =⋅⋅−⋅+⋅⇒

Example 4


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l=

1 m

a a 2a

F= 4200 kN

a b c

1 2 3

Determine N in rods. I 450, a=1m.

Conditions of solution: elastic behaviour of the rods, ideally stiff slab

Example 5

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l=

1 m

a a 2a

F= 4200 kN

a b c

Rb RcRa

1 2 3

N1

N1

N2 N3

N3N2

Ra = N1, Rb = N2, Rc = N3

Determine normal forces in the rods. Cross sections are I 450.

∑ = 0Fiz

∑ = 0Fix

∑ = 0Mia

Example 5

2) Equilibrium conditions:


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Example 5

l=

1 m

a a 2a

a b c

1 2 3

2l∆ 3l∆

1l∆F= 4200 kN


Deformation of construction:

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Chosen equations:

∑ = 0Fiz ∑ = 0Mia

a2ll

a4ll 3231 ∆−∆=∆−∆

l=

1 m

a a 2a

a b c

1 2 3

2l∆ 3l∆1l∆

(3 unknowns forces N1, N2, N3, choose 2 equations from Equilibrium equations)

xz

y

(coordinate system)


Example 5


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Solution:

Equilibrium equations:

∑ = 0izF

∑ ∑ == 0ia´ia MM

Deform. condition:

a2ll

a4ll 3231 ∆−∆=∆−∆

l=

1 m

a a 2a

a´

b c

1 2 3

⇒∆+∆=∆⇒ 312 lll2

-N1 - N2 - N3 + Fd =0

2.a.N2 +4.a.N3 – a.Fd = 0F= 4200 kN

2N2l/EA=N1l/EA+N3l/EA

N1 N2 N3a

a

xz

y

results: N1 = 2450kN, N2 = 1400kN, N3 = 350kN

Example 5

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Nehomogenized bar (steel pipe filled in by concret). Determine normal stress in steel and concret. d1 = 80 mm (external diameter), d2 = 70 mm (internaldiameter). E = 210GPa, Ecm = 24GPa.

Conditions of solution: - elastic behaviour of materials, - F affects uniformly to the section

l=

0,5

m

F= 112 kN

Example 6

1) 1x internally statically indetermined

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⇒∆=∆ CS ll

cc

c

ss

s

AE

lN

AE

lN =

∑ = 0verticali,F

l=

0,5

m

R

3) Equilibrium equations:

FN

- -NO

NB F - R = 0

(F –R =0)→

R = F = -N = - NS - NC

(2 unknowns, we take one equation from equilibrium equations)

2) Deformation equation:N = No + NB

F + NS + NC = 0

3) Stresses:

C

C

S

S

S A

N

A

N == 2, σσ

results: NS= -81,65kN, NC= -30,35kN, σS=-69,231MPa, σC=-7,91MPa

Example 6

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Condition of solution:Elastic (linear) behaviour of strain-stress diagram of material and uniform affect of load to cross-section area

Example 7: Reinforced concrete column

cs NN ,

cs NNF +=

cs ll ∆=∆cc

c

ss

s

AE

lN

AE

lN

.

.

.

. =

F

concrete

steel

Statically determined and indetermined examples

Unknown variables in example:

Equilibrium equation:

Deformation equation:

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Statically indeterminate problems in torsion

Statically indeterminate problems

Condition of solution: linear elastic behaviour of material

Unknowns: ( ) ( )2,xb,x1,xa,x MM,MM

Equilibrium equations: 0MMM c,xb,xa,x =+−−∑ = :0M i,x

Deformation condition:0

I.G

l.T2

1i i,ti

ii =∑=

:0i =ϕ∑

Both fixed ended shaft

l

ba

1l 2l

cxM ,

c

bxM ,axM ,

TaxM ,

bxM ,

−+

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Statically indeterminate problems in bending

Statically indeterminate problems in bending – Fourth-order Integration

Schwedlers relations

wIEM yy ′′−= ..

wIEV yz ′′′−= ..

IV.. wIEq yz =

wy ′=ϕ

( ) ?=xw

a) Fourth-order integration of the function of load

Methods of computing of statically indeterminate be ndings beams:

b) Force method

c) Methods based on energetic principles (Elasticity and plasticity II.)

Fourth-Order Integration of differential equation (from the function of load )

( )xy qwIE =IV..

( ) zxy VCqwIE −=+=′′′ ∫ 1..

( ) yxy MCxCqwIE −=++=′′ ∫∫ 21...

( ) 32

2

1 .2

... CxCx

CqwJE xy +++=′ ∫∫∫

( ) 43

2

2

3

1 .2

.6

... CxCx

Cx

CqwIE xy ++++= ∫ ∫ ∫ ∫

4 unknowns

4321 ,,, CCCC

↓4 boundary conditions

Solution:

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Statical and deformation boundary conditions

Type of the end (boundary)

Deformation Boundary Condition

Statical Boundary Condition

a

a

a

Free end

Simply supported end

Fixed end

0≠w0≠ϕ

0=w

0≠ϕ

0=w

0=ϕ

00 =′′→= wM

00 =′′′→= wV

00 =′′→= wM

00 ≠′′′→≠ wV

00 ≠′′→≠ wM

00 ≠′′′→≠ wV

Statically indeterminate problems in bending – Fourth-order Integration

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Fourth-Order Integration of differential equation

Determine the diagrams of internal forces (V, M) at staticallyindeterminate beam. Use differential relations.

l

q

ba

Example 1

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Boundary conditions:

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l

q

ba

Equations the same as Example 1, boundary conditions different

Determine the internal forces (V, M) at statically indete rminate beam by the Fourth-order integration of differential equation.


Example 2

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Force method

Statically indeterminate problems in bending – Force method

ba

zq( )xq

ba

zq( )xq

ba

=+

byM

0=bϕ

0, ≠qbϕ

0, ≠Mbϕ

0,, =+= Mbqbb ϕϕϕ→

Superposition

Superposition

Deformation conditions:Principle of the Force method:

Designate one of the reactions as redundant and eliminate it The redundant reaction is then treated as an unknown load that together with other loads must produce deformations that are compatible with the original supportsThe deflection or angular rotation at the point where the support has been eliminated is obtained by computing separately - the deformations caused by the given loads and by the redundant reactionResults will be obtained by superposition

•

•

•

•

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Example 1 – Force method

Determine the diagrams of internal forces (V, M) at staticallyindeterminate beam. Use the Force method (method of superposition).

l

q

ba

Deformation condition:

wa,q + wa,Ra = 0

l

q

ba

Ra

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wa,q + wa,Ra = 0

Determine other reactions from equilibrium conditions and constructthe diagrams of internal forces.

EI.8

l.qw

4

q,a =

EI.3

l.Rw

3a

Ra,a

/−=


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ϕb = 0 b

a

ϕb,M Mb

q

ϕb,q

q


ϕb,q + ϕb,M = 0

Determine the internal forces (V, M) at statically indete rminate beam by the Force method. (redundant reaction is M a)


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ϕb,q

q

EI.24

l.q 3

q,b

−=ϕ

EI.3

l.Mb

M,b =ϕϕb,M

M

Other reactions from equilibrium conditions



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