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
1) 1x statically indetermined in the axial task
3) Deformation condition:
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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
Determine stress in the rods, if both are from I 140.
Conditions of solution: elastic behaviour of the rods, stiff behaviour of the beam.
2) Equilibrium equations :
1) 1x statically indetermined
3) Deformation condition:
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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
+=⇒
Determine stress in the rods, if both are from I 140.
05,37q7N3N 21 =⋅⋅−⋅+⋅⇒
Example 4
3) Deformation condition:
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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:
1) 1x statically indetermined in the axial task
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Example 5
l=
1 m
a a 2a
a b c
1 2 3
2l∆ 3l∆
1l∆F= 4200 kN
Determine normal forces in the rods. Cross sections are I 450.
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)
Determine normal forces in the rods. Cross sections are I 450.
Example 5
3) Deformation condition:
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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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Fourth-Order Integration of differential equation
Boundary conditions:
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Fourth-Order Integration of differential equation
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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.
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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Deformation condition:
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
/−=
Example 1 – Force method
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ϕb = 0 b
a
ϕb,M Mb
q
ϕb,q
q
Deformation condition:
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)
Example 2 – Force method
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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
Deformation condition:
Example 2 – Force method
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