Slender Structures Load carrying principles - TU Delft
Transcript of Slender Structures Load carrying principles - TU Delft
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Hans Welleman 1
Slender Structures
Load carrying principles
Basic systems:
• Serial
• Parallel
v2019-1
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Content (preliminary schedule)
Basic cases
– Extension, shear, torsion, cable
– Bending (Euler-Bernoulli)
Combined systems
- Parallel systems
- Special system – Bending (Timoshenko)
Continuously Elastic Supported (basic) Cases
Cable revisit and Arches
Matrix Method
Hans Welleman 2
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Learning objectives
Extend the technique for basic models to systems
Find the ODE for a specific system and the boundary
conditions for the specific application
Solve the more advanced ODE’s (by hand and
MAPLE)
Investigate consequences/limitations of the model
and check results with limit cases
Hans Welleman 3
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Basic Cases
Second order DE
Extension
Shear
Torsion
Cable
Hans Welleman 4
Fourth order DE
Bending
2
2
2
2
2
2
2
2
d
d
d
d
d
d
d
d
xt
uEA q
x
wk q
x
GI mx
zH q
x
ϕ
− =
− =
− =
− =
4
4
d
d
wEI q
x=
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Model
(ordinary) Differential Equation – (O)DE
– Boundary conditions
– Matching conditions
Hans Welleman 5
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Combined Spring Systems (1 dof)
Serial springs
Hans Welleman 6
1 2
1 1 2 2
1 21 2
; ;
N N F
k u k u F
F Fu u
k k
= =
∆ = ∆ =
∆ = ∆ =
1 21 2
1 2
1 2
1 2
1 1 1
e
e
F Fu u u
k k
u F Fk k k
k kk
k k
= ∆ + ∆ = +
= + =
=+
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Combined Spring System (1 dof)
Parallel springs
Hans Welleman 7
1 2
1 2
u u u
N N F
∆ = ∆ =
+ =
( )1 1 2 2
1 2
1 2
e
e
k u k u F
k k u F
k u F
k k k
∆ + ∆ =
+ =
=
= +
constrained rotation
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Result
Serial
– Load bearing capacity is equal to weakest element (bothelements take the same load)
– Total deformation is summation of each element
– Equivalent stiffness :
Parallel
– Total load carrying capacity is summation of each element
– Both elements have the same deformation
– Equivalent stiffness :
Hans Welleman 8
1 2
1 1 1
ek k k= +
1 2ek k k= +
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Slightly more advanced ….
1 22
1
1
force equilibrium
N N F bN F
moment equilibrium l
N l Fb
+ = = −
=
2 1 1 21 2
1 2
2 1 2 1 2
1 2
2 1 2 1 2 1 22 2
1 2 1 2
22 21 2
2 2 2 22 1 1 2
or
( )
( )
1with : e
e
u u u uu u a u u b
l l
Fk l b kFb
k lk k lk kFbu a
lk k l
Fblk Fk a l b Fabk k al k bl k ab k abu F
l k k l k k
k k la bu F F k
kl k l k a k b k
∆ − ∆ ∆ − ∆ = ∆ + = ∆ +
−−
= +
+ − − + − −= =
= + = = +
Hans Welleman 9
( ) ( )
1 21
1 1 1 2
122
2 2 1 2
N FbkFbu
k lk lk k
F l b Fk l bNu
k lk lk k
∆ = = =
− −∆ = = =
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Non-dimensional relations1
Hans Welleman 10
Study this from the notes !
1T.J. Folkerts, A more general form for parallel springs, American Journal of Physics, 70(5):493-494, 2002
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Application
Hans Welleman 11
b = spring length in deformed state
l = additional cables (no stress)
a = chord which will be cut
What will happen to the block depending upon the values of a,b
and l.
Source : Johan Blaauwendraad, BD magazine 2, nov 1998.
neglect unloaded spring length
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Combine basic cases
Parallel system
– Bending and cable
– Bending and shear
Special “serial” system
– Bending and shear, Timoshenko Beam
Hans Welleman 12
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System with cable + bending
Same deflection, therefore P-system
Capacity is sum of elements
Hans Welleman 13
2
2
d
dcable
wH q
x− =
4
4
d
dbending
wEI q
x=
4 2
4 2
d d
d dbending cable
w wEI H q q q
x x− = + =
+
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A simple example
Hans Welleman 14
Influence of cable force H on deflection at midspan
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General solution ?
Hans Welleman 15
4 2
4 2
4 22 2
4 2
d d( ) ( ) sin
d d
sind d
with :d d
b c o
o
w w xEI H q x q x q
x x l
xq
w w Hl
x x EI EI
π
π
α α
− = + =
− = =
wh(x) = ………………………………………
wp(x) = ………………………………………
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reduction factor
Result
only bending
Hans Welleman 16
4
2 4
2
1sin
1
oq l xw
Hl EI l
EI
π
π
π
= ⋅
+
4
4sino
bending only
q l xw
EI l
π
π− =
reduction
factor
2
2
H
EI
l
π
2
2
1
1bending onlyw w
Hl
EIπ
−= ⋅
+
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Parallel system cable + bending
Suspension Bridge
Hans Welleman 17
Clifton Suspension Bridge (1864) crossing the Avon at Bristol.
Designed by Isambard Kingdom Brunel [source: The Telegraph.co.uk]
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Model
Hans Welleman 18
Same deflection, therefore P-system
Capacity is sum of elements
Assumption
Self weight p is taken by cable
only during construction
After completion total load p
and q is taken by the system
2
2
d
d
zH p
x− =
4 2
4 2
( )d w d z wEI H q p
dx dx
+− = +
4 2
4 2
d w d wEI H q
dx dx− =
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Some Math
Hans Welleman 19
4 2
4 2
d w d wEI H q
dx dx− =
General solution
Homogeneous solution
Particular solution, assume constant q
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I found as an answer
wh(x) = ………………………………………
wp(x) = ………………………………………
Hans Welleman 20
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Examples
Wheel load on cables with bending stiffness,
Constrained bending with tension,
Offshore risers,
High rise buildings
Hans Welleman 21
1A.L. Bouma, Mechanica van Constructies, Elasto-static van slanke structuren, VSSD, ISBN 9789040712784, 2000
1
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Wheel load
Hans Welleman 22
Model and boundary conditions?
Influence of H on deflection?
Influence of dimension of wheel?
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Constrained bending with tension
Hans Welleman 23
Model and boundary conditions?
Influence of H on deflection?
Influence of H on moment ditribution?
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Riser
Offshore, pipe from well
Deep water
In tidal water with currents
Hans Welleman 24
6
11 2
3
300 m
12 10 N
3.6 m
45 mm
2 10 N/m
1 m/s (1.9 knots)
1.8 10 N/m
l
H
Diam
thickness
E
current
q
=
= ×
=
=
= ×
=
= ×
334.4 10 N/m (submerged tube weight)p = ×
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Model
Assume constant H
Define boundary conditions
Find deformed position h answer 6.75 m
What if H is not constant …
Let’s use MAPLE ….
Hans Welleman 25
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Result
Hans Welleman 26
linear H
constant H
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High Rise
Shear frame combined with bending element
(elevator or staircase tower)
Hans Welleman 27
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Model?
Assume rigid links
Same deflection, therefore P-system
Summation of capacity
Hans Welleman 28
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Solve this problem
ODE
Boundary Conditions
Solution and interpretation
Let’s use MAPLE ….
Hans Welleman 29
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Displacement w(x)
Hans Welleman 30
primarily bending
shear
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Bending
moment M(x)
Hans Welleman 31
total moment
moment in bending element
change of sign!!
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Shear V(x)
Hans Welleman 32
total shear Vtot
VbVs
Help!!
Remarkable !
??
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Results
Fight between shear and bending
Concentrated force in the top link
All shear load to foundation of the tower
Change of sign in moment distribution
Hans Welleman 33
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Beams with bending and shear
deformation, Timoshenko beam
Hans Welleman 34
Displacement w(x) is result of:
Deformation due to shear
Deformation due to bending
Therefore “looks like” a serial system
'
'
shear
bending
w
w
γ
ϕ
=
= −
+d
d
w
xγ ϕ= −
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Combined system
Hans Welleman 35
Constitutive relation
Kinematic relation
Equilibrium
d(1)
d
d
d
w
x
x
γ ϕ
ϕκ
= +
=
d
d
d
d
eff eff
wV GA GA
x
M EI EIx
γ ϕ
ϕκ
= = +
= =
d
d
d
d
Vq
x
MV
x
= −
=
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Static determinate beams
Hans Welleman 36
2
2
2
2
d
d
d d d 1 d
d d d d
d
d
with: eff
w
x
w V M
x x x k x EI
w q M
x k EI
k GA
γ ϕ
γ ϕ
= − ⇔
= − = −
= − −
=
Rewrite:- use KR and CR for V
- use vertical EQ
k Euler Bernoulli beam
EI Shear beam
→ ∞
→ ∞
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Example
Hans Welleman 37
2
2
d
d
w q M
x k EI= − −
Moment distribution ?2
2
d( )
d
Mq x
x= −
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Result
This makes sense ….
Hans Welleman 38
22
2 2
2 4
2 4
sin sind
d
sin sin
o o
o o
x xq q l
w l l
kx EI
q l q lx xw
l lk EI
π π
π
π π
π π
= − −
= +
shear bending
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Check this out
Hans Welleman 39
use MAPLE !
What about static
indeterminate
systems ??
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Take a closer look …
Hans Welleman 40
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Lets check system again ..
Hans Welleman 41
( , ) ( )
( , ) ( )
u x z z x
w x z w x
ϕ=
=
displacements and shear deformation :
very nice !
theory of
elasticity
1 12 2
d d d2 2
d d dxz
u w w
z x xγ ε ϕ
= = + = +
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Differential equations
Hans Welleman 42
2
2
d dand
d d
d d(2)
d d
eff eff
eff
V wq V GA GA
x x
wGA q
x x
γ ϕ
ϕ
= − = = +
+ = −
2
2
d
d dand
d d
d
d d0 (1)
dd
eff eff
eff
wV GA GA
M xV
xM EI
x
wEI GA
xx
γ ϕ
ϕ
ϕϕ
= = + =
=
− + =
unknown displacement
fields:
w(x) and ϕ(x)
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System of coupled DE
Hans Welleman 43
2
2
2
2
d d0 (1)
d d
d d(2)
d d
eff
eff
wEI GA
x x
wGA q
x x
ϕϕ
ϕ
− + =
+ = −
What about the
boundary conditions ?
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Example
Hans Welleman 44
l = 8 m; q = 10 kN/m
GAeff = 500..50000 kN/m
EI = 10000 kNm2
Conclusion?
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MAPLE
Hans Welleman 45
> restart;
Aug 2017, © Hans Welleman
Timoshenko beam, shear and bending, example during lecture
> DV1:=EI*diff(phi(x),x$2)-GA*(diff(w(x),x)+phi(x))=0;
> DV2:=GA*(diff(w(x),x$2)+diff(phi(x),x))=-q;
> sol1:=dsolve({DV1,DV2},{w(x),phi(x)}): assign(sol1):
> w:=(w(x)); phi:=(phi(x));
standard relations for rotation, shear deformation and curvature:
> Gamma:=diff(w,x)+phi: kappa:=diff(phi,x):
> alpha_bending:=phi: alpha_shear:=diff(w,x):
sectional forces:
> V:=GA*Gamma: M:=EI*kappa:
boundary condition, clamped at the left, simply supported at the right:
> x:=0: eq1:=w=0: eq2:=phi=0:
> x:=L: eq3:=w=0: eq4:=M=0:
> sol2:=solve({eq1,eq2,eq3,eq4},{_C1,_C2,_C3,_C4}): assign(sol2); x:='x':
> L:=10: q:=8: EI:=10000: GA:=500:
> plot(-w,x=0..L,title="deflection w(x)");
> plot(-M,x=0..L,title="M");
> x:=0: evalf(M); evalf((-1/8)*q*L^2);
> x:='x': plot([Gamma,alpha_shear,alpha_bending],x=0..L,title="gamma,
dw/dx_shear,
dw/dx_bending",legend=["gamma","dw/dx_shear","dw/dx_bending"]);
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Numerical verification
Hans Welleman 46
COORDINATES 101
GENE 101 NODE 1 X 0.0 STEP 0.1 Y 0.0 STEP 0.0 Z 0.0
STEP 0.0
ELEMENTS 100
GENE 100 BEAM 1 NNODE 2 1 STEP 1 2 STEP 1
BEAM 1
1 E 2.0 IZ 5000 AX 100 AY 500 rho 0 NU 0.0 FREE
ELLOAD 100
GENE 100 STARTELE 1 STEP 1 DIRE YL FIXED Q 8
SUPPORTS 4
NODE 1 X FIXED
NODE 1 Z FIXED
NODE 1 PHIY FIXED
NODE 101 Z FIXED
EXECUTE 1
STATIC LINEAR
PRINT 4
DISPLACEMENTS ALL
SUPPORTREACTIONS ALL
LOADING ALL
STRESSES ALL
end
+-------------------------------------------------------------------+
| CALCULATION RESULTS |
+-------------------------------------------------------------------+
---------------------------------------------------------------------
DISPLACEMENTS
---------------------------------------------------------------------
dof/node-dir displacements
---------------------------------------------------------------------
NODE 51 X +0.000000e+000
NODE 51 Z +2.651042e-001
NODE 51 PHIY +9.895833e-003
---------------------------------------------------------------------
SUPPORT REACTIONS
---------------------------------------------------------------------
dof/node-dir support reactions
---------------------------------------------------------------------
NODE 1 X +0.000000e+000
NODE 1 Z -4.625000e+001
NODE 1 PHIY +6.250000e+001
NODE 101 Z -3.375000e+001
---------------------------------------------------------------------