Stability of structures - Solved examples
May 1, 2017 | Author: Andrej Markovic | Category: N/A
Short Description
Stability of structures - Solved examples (Reijo kouhia)....
Description
Stability of stru tures, solved example problems Reijo Kouhia
Pekka Marjamäki
De ember, 2005
1
Equilibrium paths, post riti al state and imperfe tions
Example 1.1
Determine all equilibrium paths of the stru ture onsisting of two rigid bars and
a linear elasti rotational spring. Investigate also the stability of all paths. P = λ4k/L.
k b
b @ @
P
L/2
Solution:
b @ b @b
L/2
Let's assume that the bars displa e by an angle
ϕ,
then in the middle pin the angle
will be 2ϕ.
b @ @
ϕ
2ϕ k b
P ````` `` `b @ b@ b
The total potential energy
Π
of the stru ture is thus:
1 Π = k(2ϕ)2 − P L(1 − cos ϕ) 2 ∂Π = 4kϕ − P L sin ϕ ∂ϕ ∂2Π = 4k − P L cos ϕ. ∂ϕ2
(1) (2)
(3)
The stru ture will be in equilibrium when the total potential energy attains its minimum, thus the rst variation of the TPE will vanish.
δΠ =
∂Π ∂Π δϕ = 0 ∀ δϕ 6= 0 ⇒ =0 ∂ϕ ∂ϕ ϕ= primary path 0 ⇒ 4k ϕ P = sek ondary path L sin ϕ
(4)
(5)
Let us rst investigate the primary path. A point on an equilibrium path is stable, if a small
hange (disturban e) in the equilibrium position will will in rease the value of
Π. Sin e the rst
variation is zero on an equilibrium path, then the se ond variation will determine the hange in the TPE. Sin e now
ϕ = 0, δ2Π =
∂2Π (δϕ)2 = 0 ∂ϕ2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
1
∂2Π = 4k − P L cos ϕ = 4k − P L ∂ϕ2 k ⇒ Pkr = 4 L
⇒
The primary equilibrium path is thus stable up to the point
(6)
(ϕ = 0, Pkr ).
Next, the stability properties of the se ondary path is investigated.
∂2Π = 4k − P L cos ϕ ∂ϕ2
(7)
P = 4kϕ/L sin ϕ to the ϕ > 0 ∀ϕ = 4k 1 − tan ϕ
Insering the equation of the se ondary path
∂ 2 Π ∂ϕ2 PII
The se ondary path is this stable for all values of
0, Pkr = 4k/L,
and the se ond variation of
Π
ϕ,
equation above, gives
ex ept the bifur ation point where
(8)
ϕ=
is zero.
The equilibrium paths are shown in the
λ − ϕ- oordinate
system in the gure below.
1.2 1 0.8
λ
0.6 0.4 0.2 0 -1
-0.8
-0.6
-0.4
-0.2
0
ϕ
0.2
0.4
0.6
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
0.8
1
2
Example 1.2
Determine all equilibrium paths of the stru ture onsisting of two rigid bars and
a linear elasti rotational spring. Investigate also the stability of all paths. Are there riti al points on the paths?
b @ @
ϕ0
k b
P ````` `` `b @ b @b
cos ϕ0 L/2 Solution:
cos ϕ0 L/2
The total potential energy expression is now
1 k[2(ϕ − ϕ0 )]2 − P L(cos ϕ0 − cos ϕ) Π = 2 ∂Π = 4k(ϕ − ϕ0 ) − P L sin ϕ ∂ϕ ∂2Π = 4k − P L cos ϕ ∂ϕ2 The stru ture will be in equilibrium when the total potential energy attains its minimum, thus the rst variation of the TPE will vanish.
δΠ =
∂Π δϕ = 0 ∀ δϕ ∂ϕ
∂Π = 0 ∂ϕ 4k(ϕ − ϕ0 ) ⇒P = L sin ϕ
⇒
An equilibrium state is stable if the se ond variation of the TPE is positive
δ2Π = ⇒
∂2Π > 0 ∂ϕ2 ⇒P <
Inserting the equilibrium equation
ondition for stability
∂2Π (δϕ)2 > 0 ∂ϕ2 4k . L cos ϕ
P = 4k(ϕ − ϕ0 )/L sin ϕ
in the expression above, gives the
ϕ − ϕ0 4k 1 − > 0 tan ϕ ϕ − ϕ0 < 1, ⇒ tan ϕ whi h is valid for all non-negative values of
ϕ. Thus this equilibrium path does not have riti al
points.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
3
2 1.8 1.6 1.4 1.2
λ
1 0.8 0.6 0.4 0.2 0 -2
-1.5
-1
-0.5
0
ϕ
0.5
1
1.5
2
In the gure above, dotted line shows the equilibrium path of the perfe t stru ture and solid line indi ates the stable path when
ϕ0 > 0.
ϕ 0 = 0,
The path in the negative part of
shown by a solid line is a omplementary path. The load parameter
λ = P/Pkr = P L/(4k).
λ
ϕ
axis
is de ned as
(9)
Noti e, that the omplementary path is not stable everywhere. Determine the unstable and stable parths of the omplementary path! Note too, that this means an existen e of a
riti al point on the omplementary path.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
4
Example 1.3
Determine all equilibrium paths starting from the unloaded state of the stru ture
onsisting of two rigid bars (length L/2) and a linear elasti rotational spring. Investigate also the stability of all paths. The perturbation load F = ǫ4k/L, where ǫ is a dimensionless (se ond) perturbation parameter.
b @ @
ϕ0
F k b?
P ````` `` `b @ b @b
cos ϕ0 L/2 Solution:
cos ϕ0 L/2
The total potential energy of the stru ture
Π
is
1 1 Π(ϕ; ϕ0 , ǫ) = k[2(ϕ − ϕ0 )]2 − P L(cos ϕ0 − cos ϕ) − F L(sin ϕ0 − sin ϕ) 2 2
(10)
A ne essary ondition of an equilibrium state is the stationarity of the TPE, thus the rst variation of the total potential energy must vanish
1 dΠ δϕ = 4k(ϕ − ϕ0 ) − P L sin ϕ + F L cos ϕ δϕ = 0 δΠ = dϕ 2
∀ δϕ 6= 0
(11)
An equilibrium path is thus de ned by
k ϕ − ϕ0 + 21 ǫ cos ϕ P =4 L sin ϕ This equation determines a unique path with respe t to
ǫ = 2ϕ0 .
not satisfy the ondition the unloaded state
Case ǫ = 2ϕ0
P = 0.
ϕ
(12)
if the perturbation parameters does
In su h a ase the stru ture is a straight bar of length
L
at
Let us examine this spe ial ase rst.
The equilibrium equation is now
dΠ = (ϕ − ϕ0 ) − P L sin ϕ + 4kϕ0 cos ϕ dϕ
(13)
= 4kϕ − P L sin ϕ + 4kϕ0 (cos ϕ − 1) = 0
(14)
and the two solutions are
ϕ=0 k ϕ + ϕ0 (cos ϕ − 1) P =4 L sin ϕ
primary path
PI ,
(15)
se ondary path
PII
(16)
An equilibrium state is stable if the se ond variation of
δ2Π =
Π:
d2 Π (δϕ)2 = (4k − P L cos ϕ − 4kϕ0 sin ϕ)(δϕ)2 2 dϕ
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
∀ δϕ 6= 0
(17)
5
is positive. Let us rst examine stability of the primary path, i.e. when
δ 2 Π|P =
d2 Π (δϕ)2 = (4k − P L)(δϕ)2 P 2 dϕ
The primary path is thus stable when
riti al load is thus
Pcr = 4k/L.
P < 4k/L
ϕ = 0,
and unstable when
thus
(18)
P > 4k/L,
and the
Let us examine wheather the riti al point is a symmetri or
asymmetri bifur ation point. The expression of the third variation of the TPE is
δ 3 Π|P = where
d3 Π (δϕ)3 dϕ3 P
(19)
d3 Π = −P L sin ϕ − 4kϕ0 cos ϕ dϕ3
(20)
At the riti al point the value of the third derivatve of the TPE is on
d3 Π = −4kϕ0 6= 0 dϕ3 kr
(21)
thus the riti al point is an asymmetri bifur ation point. The equilibrium path is drawn in gure 1.
Case ǫ 6= 2ϕ0
Let us examine stability of the equilibrium path, de ned in (12). The se ond
variation of the TPE
δ2Π =
d2 Π (δϕ)2 dϕ2
(22)
is obtained from the expression of the rst variation (11). An equilibrium state is stable if the se ond variation of the TPE is positive for all kinemati ally admissible variations
δϕ,
thus
in this single degree of freedom example it is su ient to investigate the sign of the se ond derivative of the TPE
d2 Π = 4k − P L cos ϕ − 2kǫ sin ϕ dϕ2
(23)
Let's insert the expression of the equilibrium path (12) in the expression above, gives
sin ϕ − (ϕ − ϕ0 ) cos ϕ − 21 ǫ d2 Π = 4k dϕ2 sin ϕ Let us examine the ases In the ase
ǫ > 2ϕ0
ǫ > 2ϕ0 ,
and
ǫ < 2ϕ0
(24)
separately.
the stru ture is below the horizonal line de ned by the supports
before applying the ompressive load, thus the stru ture will ontinue to displa e below the support line, thus
ϕ < 0.
Let us de ne
ǫ = 2ϕ0 + ǫ¯,
and the expression (24) gives
sin ϕ − ϕ cos ϕ − ϕ0 (1 − cos ϕ) − ǫ¯ d2 Π = 4k 2 dϕ sin ϕ Rak-54.131 Stability of stru tures exer ises / 12.12.2005
(25)
6
Sin e now
ϕ < 0
and both the nominator and denominator are negative, thus
positive, i.e. the path is stable when The ase
ǫ < 2ϕ0
δΠ
is always
ǫ > 2ϕ0 .
is more interesting. Now
ϕ>0
and the denominator of the expression
(24) is always positive but the nominator an have zero points. These roots an be solved from the trans endental equation
sin ϕ − (ϕ − ϕ0 ) cos ϕ − 21 ǫ = 0.
(26)
Sin e analyti al solution is impossible, let's try the asymptoti analysis assuming that the angles
ϕ
and
ϕ0
are small, thus
sin ϕ ≈ ϕ − 61 ϕ3 ,
cos ϕ ≈ 1 − 12 ϕ2 ,
and the expression (26) will has a form
1 3 ϕ 3
− 21 ϕ0 ϕ2 + (ϕ0 − 12 ǫ) = 0
(27)
The third order polynomial above an have both negative and positive values for positive values of
ϕ.
To show that, let us srt al ulate the minumum point
ϕ2 − ϕ0 ϕ = 0
=⇒
ϕ = ϕ0 .
The minimum value of the fun tion de ned in (27) (kun nagativity we get an inequality (let's de ne
ǫ < 2ϕ0
ϕ > 0)
and the ondition for the
ǫ = ηϕ0 )
− 13 ϕ20 + 1 − 21 η < 0 Taking the ondition
(28)
=⇒
η > 2 − 31 ϕ20
into a
ount we'll get a ondition for the perturbation parameter
ǫ = ηϕ0 : 2 > η > 2 − 31 ϕ20
i.e.
2ϕ0 > ǫ > (2 − 13 ϕ20 )ϕ0
for the existen e of a limit point on the equilibrium path. In the following gure, some equilibrium paths are shown for some values of the perturbation parameter
ǫ
To sum up, the equilibrium paths of this stru ture an have
•
a trivial equilibrium path and an asymmetri bifur ation point if
•
A stable equilibrium path without riti al points if
•
ǫ = 2ϕ0 . The
se ondary
path is de ned in equation (16).
An equilibrium path has a limit point if
ǫ > 2ϕ0
or if
(2 − 13 ϕ20 )ϕ0 / ǫ / 2ϕ0 ,
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
ǫ / (2 − 13 ϕ20 )ϕ0 .
7
2
ǫ = 2ϕ0 ǫ = 1.99ϕ0 ǫ = 1.8ϕ0 ǫ = 2.2ϕ0
1.5
λ
1
0.5
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
ϕ Figure 1: Equilibrium paths
λ = P/Pcr = P L/(4k).
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
8
4. Problem The stru ture in the problem 1 is an idealized olumn having a onstant bending stiness
k
Determine the spring oe ient
EI .
and how the riti al load will dier from the exa t beam
solution.
Solution:
The spring onstant
k
an be determined either by
•
letting the bifur ation loads to be equal for both models,
•
to make the displa ements at the middle equal under uniform load.
•
to make the displa ements at the middle equal under point load at the middle,
De e tion under a point load is
δPp =
1 F L3 48 EI
δqp =
5 qL4 . 384 EI
and for a unform load
For the spring-bar system the orresponding de e tions are
δPj
1 qL3 1 F L2 j and δq = . = 8 k 16 k
Let
δPj = δPp 6EI ⇒ kP = L δqj = δqp 12EI ⇒ kq = L The riti al load of the spring-bar system is thus
Pcr =
24EI 48EI 4k P ⇒ Pkr = and Pcrq = . 2 L L L2
One additional way to ompute
k
is to make the bending strain energies equal under a
uniform load.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
9
Example 1.4
Determine the equilibrium paths of the simple stru ture shown, onsisting of
rigid bars and elasti springs. Investigate also the stability of the equilibrium paths. Investigate espe ially ases k1 = k2 ja k1 = 5k2 . What kind of real stru tures these models imitate?
Solution:
The total potential energy expression is
Π=U +V 1 1 U = k1 L2 sin2 ϕ + k2 u2 + k2 [u − 2L(1 − cos ϕ)]2 2 2 V = −P u
(29)
The equilibrium paths an be obtained from the stationarity ondition of the TPE:
δΠ = Sin e the variations of the displa ement
∂Π ∂Π δϕ + δu = 0 ∂ϕ ∂u u
and rotation
ϕ
(30)
are arbitrary, the equilibrium paths
are obtained from equations
∂Π = k1 L2 sin ϕ cos ϕ + k2 [u − 2L(1 − cos ϕ)](−2L sin ϕ) = 0 ∂ϕ ∂Π = 2k2 u + k2 [u − 2L(1 − cos ϕ)] − P = 0 ∂u (31)
After some manipulations we get
sin ϕ[k1 L2 cos ϕ − 2Lk2 u + 4k2 L2 (1 − cos ϕ)] = 0 P 2 u= + L(1 − cos ϕ) 3k2 3 Rak-54.131 Stability of stru tures exer ises / 12.12.2005
(32) (33)
10
Equation (32) is satis ed, if
sin ϕ = 0
tai
(k1 − 4k2 )L2 cos ϕ + 4k2 L2 − 2k2 Lu = 0 k2 = k P = kL 4 + ( 32 α − 4) cos ϕ
If equation (33) is put into equation (34) and de ne
⇒
whi h is the proje tion of the se ondary path onto the
ja
k1 = αk ,
(34) we get (35)
(ϕ, P )-plane. A
ordingly from equation
(33) we get
cos ϕ = 1 +
3u P − , 2kL 2 L
whi h is substituted into (35)
⇒
P =
kL h ui 2α + (8 − 3α) , 4−α L
whi h des ribes the proje tion of the se ondary path onto the
(u, P )-plane.
The primary paths are de ned as
ϕ = and the se ondary paths
Let's investigate the
2
0 P u = 3k
P = [4 + ( 3 α − 4) cos ϕ]kL 2 h kL ui P = 2α + (8 − 3α) 4−α L
ases α = 1 ja α = 5. P = (4 − 5 cos ϕ)kL 2 α=1⇒ P = 1 kL 2 + 5 u 3 L
1.8 1.6 1.4 1.2 P kL
1 0.8 0.6 0.4 0.2 0 -0.3
-0.2
-0.1
0
ϕ
0.1
0.2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
0.3
11
4 3.5 0.333*(2+5*x) 3*x
3 2.5 P kL
2 1.5 1 0.5 0 0
0.2
0.4
0.6
0.8
1 u/L
1.2
1.4
1.6
1.8
2
We noti e, that displa ements are in reasing more rapidly on the se ondary path than in the primary path. However, the load an still be in reased over the riti al value at the bifur ation point.
(Pkr = 32 kL,
thus the se ondary path is stable. In ompressed thin plates
su h kind of behaviour an be obtained. The strong stability of the se ondary paths an be utilized also in design for some ases.
P = (4 + 7 cos ϕ)kL 2 α=5⇒ P = −kL 10 − 7 u L
8 7 6 5 P kL
4 3 2 1 0 -0.3
-0.2
-0.1
0
ϕ
0.1
0.2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
0.3
12
8 7 6 5 P kL
-10+7*x 3*x
4 3 2 1 0 0
0.5
1
1.5 u/L
2
2.5
3
In this ase the bifur ation load is mu h higher than in the preious one. However, the se ondary equilibrium path is now unstable. Shells, espe ially exhibit su h kind of unstable behaviour after bifur ation. If the post-bu kling regime is unstable, su h stru tures are imperfe tion sensitive, whi h means that the riti al load of an imperfe t stru ture is mu h lower than the theoreti al bifur ation load. Imperfe tions are due to e
entri ities, geometri al deviations et .
Example 1.5
Investigate the ee t of imperfe tions in the previous example. Draw the im-
perfe tion sensitivity diagram for the ase k1 = 5k2. Solution:
Let's determine the riti al load as a fun tion of
Now
ϕ
the imperfe tion amplitude
ϕ0 .
1 1 U = k1 L2 (sin ϕ − sin ϕ0 )2 + k2 u2 + [u − 2L(cos ϕ0 − cos ϕ)]2 2 2
and
∂Π = k1 L2 (sin ϕ − sin ϕ0 ) cos ϕ + k2 [u − 2L(cos ϕ0 − cos ϕ)](−2L sin ϕ) = 0 ∂ϕ ∂Π = 2k2 u + k2 [u − 2L(cos ϕ0 − cos ϕ)] − P = 0 ∂u Solving
u
from the equation above and substitute it into the equation below, gives
k1 sin ϕ − sin ϕ0 L − 2L(cos ϕ − cos ϕ0 ) 2k2 tan ϕ 3k1 sin ϕ − sin ϕ0 L − 8k2 L((cos ϕ − cos ϕ0 ) P = 3k2 u − 2k2 L((cos ϕ − cos ϕ0 ) = 2 tan ϕ u =
Let's program the equations into matlabiin and draw the fure (2).
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
13
λmax
as a fun tion of the imperfe tion amplitude
0.5
0.6
Figure 2: The maximum load
ϕ0
8 7 6 5 P kL
4
ǫ0 = 0 0.003 0.03 0.1
3 2 1 0
0
0.1
0.2
0.3
0.4
0.7
0.8
0.9
1
ϕ Figure 3: Equilibrium paths with dierent imperfe tion amplitude.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
14
Investigate stability properties of the stru ture below, are there limit or bifur-
Example 1.6
ation points on the paths? (EI = EA = ∞) c
Solution:
k a
L
k a
L
c c c
P
L
Let's determine the displa ements by using the following gure
ab ϕ2 b b b b b v2 b b b3 ϕ bc
ϕ1 ! a! ! ϕ 12 ! !! ! ! v1 !! ! c! ϕ0
sin ϕ0 =
v2 − v1 v2 v1 , sin ϕ12 = , sin ϕ3 = L L L
In addition
v1 v2 − v1 − arcsin L L v2 − v1 v2 = arcsin + arcsin L L
ϕ1 = ϕ0 − ϕ12 = arcsin ϕ2 = ϕ3 + ϕ12
Assuming small rotations we an approximate springs are
ϕ1 = = ϕ2 = =
arcsin x ≈ x + 1/6 x3 ,
and the rotations at the
1 1 3 2 δ1 + δ1 − δ2 − δ1 + (δ2 − δ1 ) 6 6 1 3 1 3 1 2 1 2 2δ1 − δ2 + δ1 − δ2 + δ1 δ2 − δ1 δ2 3 6 2 2 1 3 1 δ2 + δ2 + δ2 − δ1 + (δ2 − δ1 )2 6 6 1 3 1 3 1 2 1 2 2δ2 − δ1 + δ2 − δ1 − δ1 δ2 + δ1 δ2 3 6 2 2
De e tion under the load
q q p 2 2 2 ∆ = L 3 − 1 − δ1 − 1 − (δ1 − δ2 ) − 1 − δ2 3 2 2 1 3 1 4 1 4 1 3 2 2 ≈ L δ1 + δ2 − δ1 δ2 + δ1 + δ2 − δ1 δ2 + δ1 δ2 − δ1 δ2 4 4 2 4 2 Rak-54.131 Stability of stru tures exer ises / 12.12.2005
15
The total potential energy expression is
1 1 Π = Π(δ1 , δ2 ) = kϕ21 (δ1 , δ2 ) + kϕ22 (δ1 , δ2 ) − P ∆(δ1 , δ2 ) 2 2 Equilibrium paths are determined from the stationarity ondition
δΠ = 0,
whi h gives
∂Π ∂ϕ1 ∂ϕ2 ∂∆ = kϕ1 + kϕ2 −P =0 ∂δ1 ∂δ1 ∂δ1 ∂δ1
(36)
∂Π ∂ϕ1 ∂ϕ2 ∂∆ = kϕ1 + kϕ2 −P =0 ∂δ2 ∂δ2 ∂δ2 ∂δ2
(37)
in whi h
∂ϕ1 ∂δ1 ∂ϕ2 ∂δ1 ∂∆ ∂δ1 ∂ϕ1 ∂δ2 ∂ϕ2 ∂δ2 ∂∆ ∂δ2
1 = 2 + δ12 + δ22 − δ1 δ2 2 1 2 1 2 = −1 − δ1 − δ2 + δ1 δ2 2 2 3 2 3 2 1 3 3 = L 2δ1 − δ2 + δ1 − δ1 δ2 + δ1 δ2 − δ2 2 2 2 1 2 1 2 = −1 − δ1 − δ2 + δ1 δ2 2 2 1 2 = 2 + δ2 − δ1 δ2 + δ12 2 1 3 3 2 3 2 3 = L 2δ2 − δ1 + δ2 − δ1 + δ1 δ2 − δ1 δ2 2 2 2
Equations (36) and (37) are satis ed, if
ϕ1 = ϕ2 = ∆ = 0
i.e.
δ1 = δ2 = 0.
Let's investigate
stability of this primary path. The se ond variation of the total potential energy is
∂2Π ∂2Π ∂2Π 2 [δ(δ )] + 2 [δ(δ2 )]2 δ(δ )δ(δ ) + 1 1 2 2 ∂δ12 ∂δ ∂δ ∂δ 2 1 2 ! ∂2Π ∂ 2 Π2 δ(δ ) 1 ∂δ1 ∂δ2 ∂δ1 , = (δ(δ1 ) δ(δ2 )) ∂2Π ∂2Π δ(δ2 ) 2
δ2Π =
∂δ1 ∂δ2
in whi h the matrix if
2
δ Π > 0.
K = [∂ 2 Π/∂δi ∂δj ]
∂δ1
is the stability matrix. The path is stable, if and only
This is true if the stability matrix
K
is positive de nite, whi h means that all its
eigenvalues are positive.
δ1 = δ2 = 0 the elements of the stability matrix are 2 2 ∂ 2 ϕ2 ∂2∆ ∂ 2 ϕ1 ∂ϕ1 ∂ϕ2 ∂2Π + kϕ − P = kϕ + k + k 2 1 ∂δ12 ∂δ12 ∂δ1 ∂δ12 ∂δ1 ∂δ12
On the primary path
(38)
∂2Π = 5k − 2P L ∂δ12 ∂2Π = −4k + P L ∂δ1 ∂δ2 Rak-54.131 Stability of stru tures exer ises / 12.12.2005
16
∂2Π = 5k − 2P L ∂δ22 Denoting
P = λk/L
and solving the eigenvalues of
K.
The path is stable, when
K
is
positive de nite, i.e. all its eigenvalues are positive
K x¯ = ω x¯ ⇒ (K − ωI)¯ x = 0 ⇒ det(K − ωI) = 0 ! 5 − 2λ − ω λ−4 det =0 λ−4 5 − 2λ − ω ω1 = 1 − λ ja ω2 = 9 − 3λ λ
The zero points of the eigenvalues o
ur when path is stable when
λ < 1.
The eigenmodedes of
obtained from
λ=1⇒
3 −3
−3 3
!(
K
δ1 δ2
)
a
λ=3⇒
−1 1
=
(
0 0
)
is:
a` ``` ``` ``` c
1 −1
!(
δ1 δ2
)
The primary
⇒ δ1 = δ2
Pkr = k/L
=
(
0 0
)
aa a aa aa aa aaa
And the bu ling mode orresponding to the riti al load
c
λ = 3.
and
i.e. the bu kling modes of the stru ture are
The bu kling mode orresponding to the riti al load
c
λ=1
have values
c
c
⇒ δ1 = −δ2
Pkr = 3k/L
is:
c
c
c
Let's nally investigate the post-bifur ation paths after the bran hing point at
δ1 = δ2 = δ ⇒ ϕ1 = ϕ2 into the equation of equilibrium = 0 ⇒ k δ + 16 δ 3 2 + 21 δ 2 − 1 − P L δ − 21 δ 3 = 0 ⇒ δ k 1 + 16 δ 2 1 + 21 δ 2 − P L 1 + 21 δ 2 = 0 ⇒ δ = 0 tai PII = 1 + 61 δ 2 Lk
Substituting displa ements
∂Π ∂δ1
The same orresponding to the higher bifur ation load:
−ϕ1 )
∂Π ∂δ1
λ=3
kohdalla (δ1
(36).
(39)
= −δ2 = δ ⇒ ϕ2 =
= 0 ⇒ k 3δ + 32 δ 3 2 + 52 δ 2 − (−1 − 2δ 2 ) − P L 3δ + 29 δ 3 = 0 ⇒ δ 3k 1 + 12 δ 2 3 + 29 δ 2 − P L 3 + 29 δ 2 = 0 ⇒ δ = 0 tai PIII = 1 + 21 δ 2 3k L
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
λ = 1.
(40)
17
Let's draw the paths
4 3.5 3 2.5
λ
PII PIII
2 1.5 1 0.5 0 -0.4
-0.2
0
0.2
0.4
δ If we want to investigate stability properties of the paths
PII
and
PIII
we have to substi-
tute the equations of the paths (39) and (40) into the expression of the se ond variation of the TPE (38). For path
∂ 2 Π ∂δ12
PII
∂ Π ∂δ12 2
PIII
δ
it is valid (δ1
= δ2 )
2 1 2 1 2 3 2 1 3 1 3 2 2+ δ + k δ + δ · 0 + k(−1) − k 1 + δ = k δ+ δ δ+k 2+ δ 6 2 6 6 2 1 7 = k 3 + δ2 + δ4 > 0 ∀ δ 6 6
For the path
when
PII
PIII (δ1 = −δ2 , ϕ1 = −ϕ2 ) 2 5 2 3 3 3 3 − k 3δ + δ (−2δ) + k(−1 − 2δ 2 )2 = k 3δ + δ 3δ + k 2 + δ 2 2 2 1 15 −3k 1 + δ 2 2 + δ2 2 2 3 7 = k −1 + δ 2 + δ 4 < 0, 2 2
is su iently small. Therefore the path
PII
is stable and the path
PIII
is unstable near
the bifur ation point.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
18
Example 1.7
Determine the riti al load of the rigid frame suppotred by two linearly elasti
translational springs. Investigate also stability of the paths. It is assumed that the point C is not moving horizontally.
P
a D-?
C
A
a-
6
L
θ
B
?
?v
Solution:
The total potential energy expression is
1 1 2 k∆A + k∆2B − P ∆D 2 2 = k(v 2 + a2 sin2 θ) − P [v + L(1 − cos θ)] 1 = ka2 (u2 + sin2 θ) − P u + (1 − cos θ) , α
Π(v, θ) =
in whi h
v = au
and
a = αL.
P = λka, we get the form Π 1 2 2 ˜ = Π = u + sin θ − λ u + (1 − cos θ) ka2 α By de ning
The equilibrium equations are
˜ λ ∂Π = 2u − λ = 0 ⇒ u = ∂u 2 ˜ 1 ∂Π = 2 sin θ cos θ − λ sin θ = 0 ∂θ α 1 = sin θ 2 cos θ − λ = 0 α ⇒ Thus the primary path is de ned as
Let's substitute the
(
sin θ = 0 ⇒ θ = 0 λ = 2α cos θ
u = λ/2
ja
θ = 0.
Stability of the primary path
˜ ˜ ˜ ∂2Π ∂2Π ∂2Π λ = 2, = 0, = 2 cos 2θ − cos θ ∂u2 ∂u∂θ ∂θ2 α expressions of the primary path u = λ/2 and θ = 0,
matrix:
K=
"
2
0
0 2−
λ α
into the stability
#
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
19
The riti al value of the load parameter
λkr
an be obtained from the ondition det(K)
2
λkr = 2αka = 2α kL. λ = 2α cos θ: " # " # 2 0 1 0 K= =2 0 2 cos 2θ − 2 cos2 θ 0 − sin2 θ
On the se ondary equilibrium path
The se ondary path is unstable sin e
u = λ/2
=0⇒
ja
2 ˜ K2 2 = ∂ 2 Π/∂θ = − sin2 θ ≤ 0 ∀ θ.
2
1.5
λ/α
1
0.5
0 -0.4
-0.2
0
0.2
0.4
θ
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
20
2
Continuous models, beams et .
Example 2.1
Derive the expression of urvature for a plane beam using (a) the Lagrangian
and (b) the Eulerian approa h. Solution:
The dieren e between the Lagrangian and Eulerian approa hes is the meaning of
the independent variable
x. In the Lagrangian approa h the oordinate is atta hed to a material
point. The dispa ement at point
x is
the displa ement of the point a
upying the position
x
at
the initial undeformed on guration. In the Eulerian approa h the oordinate is referring only to a spatial point
x.
da = dx
x, u
ϕ
?
dv
y, v
da u
Lagrange:
-
u + du
It is seen from the gure
dv dv = = v′ da dx ⇒ ϕ = arcsin v ′
sin ϕ =
sin e the urvature is
κ = 1/R = ϕ′
we get
1 v ′′ , = ϕ′ = p R 1 − (v ′ )2
d arcsin x 1 =√ dx 1 − x2
da
-
x, u
ϕ
?
dv
y, v
da dx
Euler:
From the gure
da =
√
p dv 2 + dx2 = dx (v ′ )2 + 1
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
21
dv = v′ dx ⇒ ϕ = arctan v ′
tan ϕ =
we obtain for the urvature
1 ∂ϕ ∂ϕ κ= not = !! R ∂a ∂x d arctan x 1 ∂ dv 1 where = = 1 + v ′2 ∂a dx dx 1 + x2 2 √ 1 1 dv √ = (where da = dx 1 + v ′2 ) ′2 2 1 + v dx 1 + v ′2 v ′′ v ′′ √ = = (1 + v ′2 ) 1 + v ′2 (1 + v ′2 )3/2 Note! When the higher order terms are negle ted we get the same result for both approa hes:
κ = v ′′
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
22
Example 2.2
Determine Pcr starting from the dierential equation.
2 EI
e
EI
x
L/2
e
P
-
e e
L/2
Solution: In part 1 1
(4)
v1 + k 2 v1′′ = 0,
(4) 2 v2
where
k 2 = P/2EI .
(see problem ??)
=0
BC : v1 − L2 = v1′ − L2 = v2 v1 (0) = v2 (0)
L 2
= v2′
L 2
=0
v1′ (0) = v2′ (0)
M -1
M2 X z XXX X XP X XXX XX XQ2 Q1 X
P X
M1 (0) = M2 (0) Q1 (0) = Q2 (0) + P v2′ (0) Solutions for the homogenious equations are
v1 = C1 sin kx + C2 cos kx + C3 x + C4 v1′ = C1 k cos kx − C2 k sin kx + C3
v1′′ = −C1 k 2 sin kx − C2 k 2 cos kx
v1′′′ = −C1 k 3 cos kx + C2 k 3 sin kx v2 = C5 x3 + C6 x2 + C7 x + C8 v2′ = 3C5 x2 + 2C6 x + C7 v2′′ = 6C5 x + 2C6 v2′′′ = 6C5
Taking the boundary onditions into a
ount
Q1 (0) = Q2 (0) + P v2′ (0) −2EIv1′′′ (0) = −EIv2′′′ (0) + P v2′ (0) 2C1 k 3 = −6C5 + 2k 2 C7 1 1 C5 = − k 3 C1 + k 2 C7 3 3
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
23
M1 (0) = M2 (0) −2EIv1′′ (0) = −EIv2′′ (0)
2C2 k 2 = −2C6 ⇒ C6 = −k 2 C2 v1′ (0) = v2′ (0)
1 1 1 C1 k + C3 = C7 ⇒ C5 = − k 3 C1 + k 2 (C1 k + C3 ) = k 2 C3 3 3 3 v1 (0) = v2 (0) C2 + C4 = C8 kL kL L L = 0 ⇒ C4 = C1 sin − C2 cos + C3 v1 − 2 2 2 2 L kL kL v1′ − = 0 ⇒ C3 = −k(C1 cos + C2 sin ) 2 2 2 3 2 1 2 L L L L 2 = 0 ⇒ k C3 − k C2 + (C1 k + C3 ) + C2 + C4 = 0 v2 2 3 2 2 2 kL 1 kL kL ⇒ 1 + (kL)2 + sin C1 − kL cos 2 2 24 2 1 kL kL 1 2 2 1 + (kL) C2 = 0 − kL sin + 1 − (kL) − cos 4 2 2 24 2 L L L ′ 2 v2 = 0 ⇒ k C3 − 2k 2 C2 + C1 k + C3 = 0 2 2 2 kL 1 kL 1 2 2 kC1 + −kL − 1 + (kL) sin C2 = 0 ⇒ 1 − 1 + (kL) cos 4 2 4 2
⇒
"
a(kL) b(kL) c(kL) d(kL)
#(
C1 C2
)
=
(
0 0
det = 0 ⇒ kL ≈ 7.55 ⇒ Pkr = 114
) EI L2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
24
Example 2.3
Determine Pcr starting from the dierential equation. (α = 2, β = 1)
α β
EI
EI
EA
EA
Sin e
Part 1: 2:
-
x
L/2
Solution:
@ @ @ @ @ @ @ @
P
L/2
β = 1 ⇒ (EA)1 = (EA)2 ⇒ P1 = P2 = P/2. (4) P/2 P = 4EI v1 + k12 v1′′ = 0 k12 = 2EI (4)
v2 − k22 v2′′ = 0 √ k22 = 2k12 ⇒ k2 = 2k1
k22 =
BC : v1 − L2 = v1′ − L2 = v2 v1 (0) = v2 (0)
L 2
P/2 EI
= v2′
=
P 2EI
L 2
=0
M -1 P 2 X X
v1′ (0) = v2′ (0)
2 z XXX X XP X XX Q XX X X X2 Q1 X z P2 X X
M
M1 (0) = M2 (0) Q1 (0) = Q2 (0) + P v2′ (0) Solutions for the homogenious dierential equations are
v1 = C1 sin kx + C2 cos kx + C3 x + C4 v1′ = C1 k cos kx − C2 k sin kx + C3
v1′′ = −C1 k 2 sin kx − C2 k 2 cos kx
v1′′′ = −C1 k 3 cos kx + C2 k 3 sin kx
v2 = C5 sinh kx + C6 cosh kx + C7 x + C8 v2′ = C5 k cosh kx − C6 k sinh kx + C7
v2′′ = −C5 k 2 sinh kx − C6 k 2 cosh kx
v2′′′ = −C5 k 3 cosh kx + C6 k 3 sinh kx Taking the boundary onditions into a
ount
Q1 (0) = Q2 (0) + P v2′ (0) −2EIv1′′′ (0) = −EIv2′′′ (0) + P v1′ (0) 2C1 k13
C5
= −C5 k23 + 4k12 (C1 k1 + C3 ) 2 1 1 = − 3 2k13 C1 + 4k12 C3 = √ C1 + √ C3 k2 2 2k1
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
25
M1 (0) = M2 (0) −2EIv1′′ (0) = −EIv2′′ (0) 2C2 k12
=
−C6 k22
⇒ C6 = −2
k1 k2
2
C2 = −C2
v1′ (0) = v2′ (0) C1 k 1 + C3
= C5 k2 + C7 ⇒ C7 = C1 k1 + C3 − C5 k2 = −C3
v1 (0) = v2 (0) C2 + C4
= C6 + C8 ⇒ C8 = C2 + C4 − C6 = 2C2 + C4
k1 L k1 L L L = 0 ⇒ C4 = C1 sin − C2 cos + C3 v1 − 2 2 2 2 L k1 L k2 L v1′ − = 0 ⇒ C3 = −k1 C1 cos + C2 sin 2 2 2 k1 L k1 L k1 L k1 L k1 L k1 L C1 − cos C2 − cos + sin ⇒ C4 = sin 2 2 2 2 2 2 L 1 kL L k L 2 √ C1 + √ C3 sinh 2 − C2 cosh 2 − C3 + 2C2 + C4 = 0 v2 = 2 2 2 2 2k1 2 2 k1 L k2 L k1 L 1 √ − √ cos sinh C1 + sin ⇒ 2 2 2 2 2 2 k2 L k2 L k1 L k1 L − √ sin C2 = 0 sinh − cosh + 2 − cos 2 2 2 2 2 k2 L 1 2 2 k1 L k1 L L ′ √ C1 − √ C1 cos = k2 cosh − √ C2 sin v2 2 2 2 2 2 2 2 k2 L k1 L k1 L −C2 k2 sinh + k1 C1 cos + k1 C2 sin =0 2 2 2 k2 k2 L k1 L k2 L 2k2 k1 L ⇒ √ cosh C1 − √ cos cosh + k1 cos 2 2 2 2 2 2 k1 L k2 L 2k2 k2 L k1 L + k1 sin C2 = 0 − k2 sinh − √ sin cosh 2 2 2 2 2
⇒
"
a(kL) b(kL) c(kL) d(kL)
#(
C1 C2
)
=
(
The riti al load is obtained from the equation det[℄
0
)
= ad − bc = 0
⇒ k1 L ≈ 10, 637 ⇒ Pkr = 4k12 If the dire tion of the load is reversed, the result is
0
EI EI = 452, 6 2 2 L L
247, 5EI/L2.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
26
Example 2.4
Derive the Euler equations of the antilever beam shown below. Assume inex-
tensible beam and small de e tions. Solve the equations and determine the eigenmodes and show that the eigenmodes are orthogonal.
EI
P
L
Solution:
The total potential energy fun tional is
1 Π(v) = 2
ZL 0
where the horizontal de e tion
EI(v ′′ )2 − P (v ′)2 dx,
∆ under the load P
an be determined as
ϕ
dx + du
The Euler equations are obtained from the stationarity ondition of the fun tional
δΠ = Π,v δv =
ZL 0
where
δv
is the variation of the de e tion, i.e. an arbitrary fun tion satisfying the homogenious
v(0) = v ′ (0) = 0.
kinemati al boundary onditions
δv
(EIv ′′ δv ′′ − P v ′ δv ′ )dx = 0,
After integration by parts we get the term
as a ommon fa tor inside the integral
L L ZL ZL ′ ′′ ′ ′′ ′ ′ δΠ = EIv δv − (EIv ) δv dx − P v δv + δvdx 0
0
0
0
L L L ZL ′ ′′ ′ ′′ ′ = EIv δv − (EIv ) δv − P v δv + [(EIv ′′ )′′ + (P v ′ )′ ] δvdx 0
At the lower limit
0
δv(0) = δv ′ (0) = 0,
0
0
and taking into a
ount the de nitions of the moment
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
27
dv
and shear for e:
M = −EIv ′′
sekä
Q = −(EIv ′′ )′ ,
′
we get
ZL
′
δΠ = −M(L)δv (L) + [Q(L) − P v (L)]δv(L) +
[(EIv ′′ )′′ + (P v ′ )′ ] δvdx = 0,
0
sin e
δv
is arbitrary fun tion satisfying the boundary onditions
(EIv ′′ )′′ + (P v ′)′ = 0 M(L)
=0
Q(L) − P v ′ (L) = 0
lowing equations have to be satis ed
v(0)
=0
v ′ (0) = 0 If the bending stiness
EI
v(0) = v ′ (0) = 0, thus the folx ∈ (0, L) (Euleri equation) natural
)
boundary onditions
essential
)
and the ompressive for e
boundary onditions
P
are onstants in the domain, we
get a homogeneous dierential equation with onstant oe ients
EIv (4) + P v ′′ = 0 ⇒ EIv ′′ + P v = Cx + D, (C, D constants) ⇒ v = A sin kx + B cos kx + Cx + D, k =
r
P EI
The derivatives are
v ′ = Ak cos kx − Bk sin kx + C
v ′′ = −Ak 2 sin kx − Bk 2 cos kx
v ′′′ = −Ak 3 cos kx + Bk 3 sin kx
v(0) = 0 ⇒ B + D = 0
v ′ (0) = 0 ⇒ Ak + C = 0
v ′′ (L) = 0 ⇒ A sin kL + B cos kL = 0 −EIv ′′′ (L)−P v ′ (L) = 0 ⇒ −EI(−Ak 3 cos kL+Bk 3 sin kL)−P (Ak cos kL−Bk sin kL+C) = 0 λ EI ⇒ k= 2 L L A = 0 ⇒ C = 0 ⇒ B cos kL = 0 ⇒ B = 0 P =λ
It follows from equation (41) that If
B=0 ⇒ v≡0
(41)
√
or
cos kL = 0.
it yields a trivial solution, hen e we should have
cos kL = 0 ⇒ kL =
π + nπ, n = 0, 1, 2, ... 2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
28
⇒ λn = and the lowest bu kling load is
λ0 =
π 2
π 2
+ nπ
2
π 2 EI 4 L2
⇒ Pcr =
2
λn
The eigenmode orresponding to the eigenvalue
,
is
vn = B(cos kn x − 1), kn =
1 π + nπ L 2
It was asked to give the normalized eigenmodes. For that we should de ne how this normalization should be done. It is usual to use the energy norm
ZL
||vn ||2E =
EI(vn′′ )2 dx.
0
The energy orthogonality thus means
ZL 0
Lets normalize the eigenmodes
[E1 ] =
√
vn
′′ EIvn′′ vm dx = 0, kun n 6= m. su h, that
Nm.
||vn ||E = E1 ,
vn′′ = −Bkn2 cos kn x ⇒
E12
=
EIB 2 kn4
ZL
E1
where
is the energy unit and
cos2 kn xdx
0
Lets hange variables su h, that
1 y = kn x, dx = dy rajat kn
(
x=0
⇒ y=0
x=L ⇒ y=
π 2
+ nπ
π +nπ 2
⇒
E12
=
EIB 2 kn3
Z
2
cos ydy =
1 EIB 2 kn3
0
2
2E12 2E12 = ⇒ B2 = 4 EIkn3 π2 + nπ EI π16 (1 + 2n)4 √ 3/2 4 2L E1 √ ⇒ B= π 2 (1 + 2n)2 EI
π 2
+ nπ
The energy orthonormal eigenfun tions are thus
√ 4 2 E1 L3/2 √ vn (x) = bn (cos kn x − 1), Bn = 2 π (1 + 2n)2 EI Rak-54.131 Stability of stru tures exer ises / 12.12.2005
29
Orthogonality:
ZL 0
′′ EIvn′′ vm dx = 0, kun n 6= m.
ZL 0
′′ vn′′ vm dx
=
ZL 0
hπ xi xi cos (1 + 2m) cos (1 + 2n) 2 L 2 L
2L = π
ZL
hπ
πx 2L merk. y = , dx = dy 2L π
cos[(1 + 2n)y] cos[(1 + 2m)y]dy = 0, kun n 6= m.
0
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
30
Example 2.5
Determine the maximum def etion and maximum monents at supports and in
span as a fun tion of the ompressive for e P for the beam shown below. q
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? e e
EI
Solution:
e e
P
L
The dierential equation for the beam- olumn is
v (4) + k 2 v ′′ =
q P , where k 2 = EI EI
The solution is
v = C1 sin kx + C2 cos kx + C3 x + C4 + Ax2 , where A =
q q = . 2 2EIk 2P
Let's hoose the zero o-ordinate at the midspan. From the boundary onditions we get
v ′ (0) = 0 ⇒ C1 k + C3 = 0
v′′′ (0) = 0 ⇒ C1 = 0 ⇒ C3 = 0 L kL qL v′ = 0 ⇒ −C2 k sin + =0 2 2 2P qL ⇒ C2 = 2kP sin kL 2 kL L2 L = 0 ⇒ C2 cos + C4 + A =0 v ± 2 2 4 qL2 qL − ⇒ C4 = − 8P 2kP tan kL 2
Denoting
qL kL kL q 2 kL qL cos kx − cos + + sin x ⇒ v(x) = kL kL 2 4 2 2P 2kP sin 2 2kP sin 2 √ P = λ EI/L2 , thus kL = λ. √ ! λ qL qL2 √ ⇒ v(0) = 1 − cos − 2 8P 2kP sin 2λ ! qL2 kL cos kx q ′′ 2 EI = −1 ⇒ M(x) = −EIv = C2 k cos kx − P λ 2 sin kL 2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
31
The bending moment at the lamped support
x=±
L 2
√
2
Mt =
qL λ
λ
2 tan
√
λ 2
−1
!
The bending moment in the midspan
qL2 (x = 0) Mk = λ
√
λ
2 sin
√ λ 2
−1
!
4
P/Pkr
λ
Mt /qL2
Mk /qL2
v(0)/ qL EI
0
0
-0.0833
0.0417
0.0026
0.5
2π 2
-0.1363
0.0908
0.0052
2
-0.2390
0.1911
0.0103
-0.5439
0.4944
0.0257
0.75
3π
0.9
3.6π 2
0.6 0.4
Mt Mk
0.2 M qL2
0 -0.2 -0.4 -0.6 0
0.1
0.2
0.3
0.4
0.5
P/Pcr
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
0.6
0.7
0.8
0.9
32
Example 2.6
Determine the bending moment distribution at the load levels P/PE = 0.25, 0.50
and 0.75, where PE is the riti al load of the bu kling problem. Determine also the expressions of the support moments at both ends and the bending moment in the midspan as a fun tion of the ompressive for e. F
2 EI
e
P x
L/2
e
EI
?
-
e e
L/2
Solution: Osalla 1 2
(4)
v1 + k 2 v1′′ = 0,
missä
k2 =
P 2EI
(4)
v2 = 0
BC : v1 − L2 = v1′ − L2 = 0 v2 L2 = v2′ L2 = 0 v1 (0) = v2 (0)
v1′ (0) = v2′ (0)
M -1
F
M2 X z XXX ? X XP X XXX XX XQ2 Q1 X
P X
M1 (0) = M2 (0) Q1 (0) = Q2 (0) + P v2′ (0) + F Solution for the homogeneous dierential equations are:
v1 = C1 sin kx + C2 cos kx + C3 x + C4 v1′ = C1 k cos kx − C2 k sin kx + C3
v1′′ = −C1 k 2 sin kx − C2 k 2 cos kx
v1′′′ = −C1 k 3 cos kx + C2 k 3 sin kx v2 = C5 x3 + C6 x2 + C7 x + C8 v2′ = 3C5 x2 + 2C6 x + C7 v2′′ = 6C5 x + 2C6 v2′′′ = 6C5
Taking the boundary onditions into a
ount
Q1 (0) = Q2 (0) + P v2′ (0) Rak-54.131 Stability of stru tures exer ises / 12.12.2005
33
−2EIv1′′′ (0) = −EIv2′′′ (0) + P v2′ (0) + F F 2C1 k 3 = −6C5 + 2k 2 C7 + EI 1 3 1 2 F C5 = − k C1 + k C7 + 3 3 6EI M1 (0) = M2 (0) −2EIv1′′ (0) = −EIv2′′ (0)
2C2 k 2 = −2C6 ⇒ C6 = −k 2 C2 v1′ (0) = v2′ (0)
1 F 1 F 1 = k 2 C3 + C1 k + C3 = C7 ⇒ C5 = − k 3 C1 + k 2 (C1 k + C3 ) + 3 3 6EI 3 6EI v1 (0) = v2 (0) C2 + C4 = C8 L kL kL L v1 − = 0 ⇒ C4 = C1 sin − C2 cos + C3 2 2 2 2 kL kL L = 0 ⇒ C3 = −k(C1 cos + C2 sin ) v1′ − 2 2 2 3 2 L 1 2 L F L L 2 v2 = 0⇒ k C3 + − k C2 + (C1 k + C3 ) + C2 + C4 = 0 2 3 6EI 2 2 2 kL 1 kL kL 1 + (kL)2 + sin C1 − kL cos ⇒ 2 2 24 2 1 F L3 kL kL 1 2 2 1 + (kL) C2 = − − kL sin + 1 − (kL) − cos 4 2 2 24 48EI 2 L F L L v2′ = 0 ⇒ k 2 C3 + − 2k 2 C2 + C1 k + C3 = 0 2 2EI 2 2 1 F L2 kL kL 1 2 2 kC1 + −kL − 1 + (kL) sin kC2 = − ⇒ 1 − 1 + (kL) cos 4 2 4 2 8EI The expressions for the bending moments are
( The oe ients
M1 (x) = 2EIk 2 (C1 sin kx + C2 cos kx) M2 (x) = −EI(6C5 x + 2C6 )
C1
and
C2
C5
and
C6
L 2
≤x≤0 L 2
when 0 < x ≤
an be solved from the equation system below
( The oe ients
when −
3
FL [·] C1 + [·] C2 = − 48EI 2
L [·] kC1 + [·] kC2 = − F8EI
)
have already been solved as a fun tions of
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
C1
and
C2 . 34
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
35
Example 2.7
A beam with ir ular ross-se tion has an initial de e tion v0 (x) = v0 sin(πx/L).
What is the safety fa tor with respe t to the yield limit if the ompressive load has the value
P = 50 kN? The yield stress is σy = 220 MPa and the Young's modulus E = 210 GPa. The amplitude of the initial de e tion is v0 = L/1000. Determine also the resistan e and the partial safety fa tor γf a
ording to the Finnish steel design spe i ations B7. b b b @ @ bhhh h
b " " @ (" b @ b (((
P
r = 50mm
t = 5mm
L = 5.0 m
Solution:
The bending moment distribution due to the ompressive for e is
M(x) + P [v(x) + v0 (x)] = 0 ⇒ v ′′ (x) + k 2 v(x) = −k 2 v0 (x), where k 2 =
P EI
(42)
Let's nd the parti ular solution of the dierential equation above.
vy (x) = A sin
πx L
Substituting the trial fun tion above into equation 42
⇒ ⇒
πx πx π2 2 = −k 2 v0 sin − 2 + k A sin L L L k 2 v0 A=− 2 k 2 − Lπ 2
The solution is the sum of the general solution of the homogeneous equation and the parti ular solution
v(x) = C1 sin kx + C2 cos kx + C3 x + C4 −
πx k 2 v0 sin 2 L k 2 − Lπ 2
Boundary onditions:
v(0) = C2 + C4 = 0 v ′′ (0) = k 2 C2 = 0
⇒ C2 = 0 ⇒ C4 = 0
v ′′ (L) = k 2 C1 sin kL = 0 ⇒ C1 = 0(∗) v(L) = C3 L = 0
At
(∗)
the solution
kL = nπ
is not valid, sin e the equation must hold on for all values of
⇒ v(x) = −
k:
πx k 2 v0 , 2 sin π L k 2 − L2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
36
and the bending moment has the expression
M(x) = −EIv ′′ (x) = − The largest bending moment is at the middle
πx EIk 2 v0 π 2 sin 2 2 2 k L −π L
(k 2 = P/EI):
P v0 π 2 L = − P L2 M 2 − π2 EI The bu kling load for an ideal straight olumn is
PE = π 2 EI/L2 ,
the bending moment an be
expressed as
P v0 L =− P M 2 −1 P E
The bending moment
M(L/2) approa hes
to in nity when
of the beam in the outmost bers are
M P = −P σ=− ± A W
1 ± A
P → PE ! The stresses
1 v0 P −1W PE
!
at the middle
(43)
Taking the ross-se tion dimensions into a
ount
A = π(502 − 452) = 1492mm2
I = π4 (504 − 454 ) = 1.688 · 106 mm4
W
k2
=
1 I 50 mm2 −3
= 1.41 · 10
= 33760mm3
, when P = 50 kN
From equation 43 we get
σ = −33.5 ± 15.4 MPa Let's solve the ompressive for e value the yield point
σy .
P,
when the outmost bers at the mid-se tion attains
From the equation 43 we get
1 A
v0 1 P −1 W P
σy = −P + E v P P 0 − 1 σy − A − W P =0 ⇒ PE ⇒ P 2 − σm A + PE + PEWAv0 P + σy PE A = 0 Substituting the dimensions, gives
2
P − 509.5P + 45945 = 0 ⇒
(
P1 = 117.1 kN P2 = 392.4 kN
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
37
Safety fa tor with respe t to the yield is thus
n=
117.1 = 2.34 50
Resistan e and partial safety fa tor a
ording to B7.
•
The resistan e of the beam depends on the hosen bu kling urve. Dieren e between the
•
Let's use the bu kling urve C
bu kling urve is in the assumed level of imperfe tion
i = ¯k = λ β = fck = NRc
From the resistan e
NRc
v0
⇒ α = 0.49 r I = 33.6 mm A r Lc fy = 1.53 iπ E ¯ k − 0.2) + λ ¯2 1 + α(λ k = 0.852 2 ¯ 2 λ k q ¯ 2 )fy = 66.99 MPa (β − β 2 − 1/λ k
A = fck = 66.99 · 1492 = 99.97 kN γm
we get the partial safety fa tor
γf = If we hoose the bu kling urve A
99.97 = 2.0 50
⇒ α = 0.21, γf =
it results in
118 = 2.36 50
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
38
Example 2.8
An elasti beam with ir ular ross se tion is loaded by a tensile for e and a
twisting moment. Determine the riti al twisting moment when the beam loses its stability. Does the beam bu kle if the for e N is ompressive?
M t --
Mt
Solution:
@ @
@ @
-
N
The equilibrium equations at the deformed state
Let's investigate the proje tions of
the
de e tion
urve
on
the
o-
ordinate planes
The bending moments aused by the normal for e are
−Ny
ja
−Nz .
The twistiong moment produ es the bending moments
Mz ′
in
aiheutuu
Vääntömomentista
taavasti and
taivutusmomentit
−My ′
(bending in
(bending vas-
xy -plane)
xz -plane).
We get the system of equations
−EIy ′′ = −Ny + Mz ′
−EIz ′′ = −Nz − My ′
Let's try the solution of the form
⇒
(
(44)
y = C1 erx , z = C2 erx
EIr 2 C1 erx = NC1 erx − MC2 rerx EIr 2 C2 erx = NC2 erx + MC1 rerx
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
(45)
39
(45)1 ⇒ C2 (EIr 2 − N) = MC1 r (45)2 ⇒ EIr 2 C1 − NC1 + Mr
MC1 r =0 EIr 2 − N
⇒ (EIr 2 − N)2 + M 2 r 2 = 0 By denoting
α2 = −r 2
(46)
we get from the solutions of the equations (46) the following equations
−EIα2 − N = Mα ⇒ α2 +
M α EI
EIα2 + N = Mα ⇒ α2 −
M α EI
The latter ase is not valid sin e
α1 , α2 > 0. α1,2 =
+
N EI
=0 (47)
+
N EI
=0
Therefore
M ± − EI
q
M 2 EI
2
N − 4 EI
(48)
Thus
y = A1 sin α1 x + B1 cos α1 x + C1 sin α2 x + D1 cos α2 x z = A2 sin α1 x + B2 cos α1 x + C2 sin α2 x + D2 cos α2 x
Substituting these expressions ba k to the equations (441 ) give
EIz ′′ = EIA2 (−α12 ) sin α1 x − EIB2 α12 cos α1 x − EIC2 α22 sin α2 x − EID2 α22 cos α2 x −Nz = −NA2 sin α1 x − NB2 cos α1 x − NC2 sin α2 x − ND2 cos α2 x
−My ′ = MB1 α1 sin α1 x − MA1 α1 cos α1 x + MD1 α2 sin α2 x − MC1 α2 cos α2 x
Sin e the equation
EIz ′′ − Nz − My ′ = 0
must hold for all the values of
x,
the following
equations must be ful lled
(−EIα12 − N)A2 + MB1 α1 = 0
(−EIα12 − N)B2 − MA1 α1 = 0
(−EIα22 − N)C2 + MD1 α2 = 0
(−EIα22 − N)D2 − MC1 α2 = 0
Taking equations (47) into a
ount we get the following relationships for the oe ients
A2 =
M B1 α1 EIα21 +N
=
M B1 α1 −M α1
= −B1
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
40
B2 = C2 = D2 =
Therefore
−M A1 α1 EIα21 +N M D1 α2 EIα22 +N −M C1 α2 EIα22 +N
−M A1 α1 −M α1 D1 α2 = M−M α2 −M C1 α2 = −M α2
=
= A1 = −D1
= C1
z = −B1 sin α1 x + A1 cos α1 x − D1 sin α2 x + C1 cos α2 x.
From the boundary onditions we get
y(0) = 0 ⇒ B1 + D1 = 0 z(0) = 0 ⇒ A1 + C1 = 0
y(L) = 0 ⇒ A1 sin α1 L + B1 cos α1 L + C1 sin α2 L + D1 cos α2 L = 0
z(L) = 0 ⇒ −B1 sin α1 L + A1 cos α1 L − D1 sin α2 L + C1 cos α2 L = 0
The riti ality ondition is the zero determinat, thus
2 − 2(cos α1 L cos α2 L + sin α1 L sin α2 L) = 2 − 2 cos(α1 L − α2 L) = 0 i.e.
(α1 − α2 )L = n2π .
On the other hand we get from equation (48) the following relationship
α1 − α2 = ±2 If
α1 = α2 ,
s
M 2EI
2
−
N EI
the solution above is not valid, therefore the trigonometri fun tions should be
repla ed by polynomials (partially). However, this is not possible. A situation where is possible, but a
ording to the equations (44) it would imply
′′
′′
y = z = 0.
N =M =0
This kind of rigid
body motion is prevented by boundary onditions. Su h kind of deformed equilibrium state is not possible when
When
α2 − α1 = 0, therefore we have to hoose (α2 − α1 )L = 2π . s 2 N M L = 2π − ⇒ ±2 2EI EI 2 π 2 M N ⇒ = − 2EI EI L r π 2 N + ⇒ Mcr = ±2EI EI L
N > 0, Mcr
always exists
N < 0, Mcr
is possible, until
N = −π 2 EI/L2
= the Euler bu kling load
This kind of phenomena is present when twisting a string. From a ertain value of the twisting moment one has to apply also a normal for e to prevent the string to be plaitened.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
41
Example 2.9
A long bolt has been pla ed inside a wide sleeve (of lengtht L) as shown in the
gure below. When starting to tighten the bolt: 1. Determine the for e when the sleeve will bu kle when assuming the rorations at the ends to be equal for the bolt and the sleeve. Use the Berry's fun tion for ompressed beam for the sleeve and the orresponding fun tions for a tensile bar for the bolt. 2. What is the value of Pcr , when Ibolt = Isleeve ? 3. What is the value of Pcr , when the normal beam oe ients for a linear bar is used for the bolt? 4. What is the result if there is no spa e between the bolt and the sleeve?
Solution: 1. Let's use the for e method and denoting the quantities related to the sleeve by a sus ript 1 and ralated to the bolt by a subs ript 2.
L L ψ1 − M21 φ1 3EI1 6EI1 L 1 = −M0 (ψ1 + ϕ1 ) 3EI1 2 L L ψ2 − M21 φ2 = M12 3EI2 6EI2 L 1 = M0 (ψ2 + ϕ2 ) 3EI2 2
sleeve : ϕ12 = M12
bolt : ϕ12
Sin e
ϕ12,sleeve = ϕ12,bolt ⇒ M0 6= 0 ⇒
L 1 1 L (ψ1 + ϕ1 ) + M0 (ψ2 + ϕ2 ) = 0 3EI1 2 3EI2 2 I1 1 1 (ψ2 + ϕ2 ) + (ψ1 + ϕ1 ) = 0 I2 2 2
M0
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
42
Substituting the Berry fun tions for the tensile and ompressed bars
I1 3 1 1 1 1 − + − I2 (kL)2 tanh(kL)2 (kL)2 (kL)2 sinh(kL) 2 1 1 1 1 3 =0 − + − + (kL)1 (kL) 1 tan(kL)1 sin(kL) 1 (kL)1 I1 (kL)1 1 1 1 1 ⇒ − + − =0 I2 (kL)2 tanh(kL)2 sinh(kL)2 sin(kL)1 tan(kL)1 ⇒ Pcr = EI1 k12
2. If
I1 = I2 ⇒ (kL)1 = (kL)2 = kL ⇒
cosh kL − 1 cos kL − 1 22.4EI = ⇒ kL ≈ 4.73 ⇒ Pcr = sinh kL sin kL L2
3. If we use the linear oe ients for the bolt, then
ψ2 = ϕ2 = 1
1 I1 3 + (ψ1 + φ1 ) = 0 I2 2 2 3 1 1 I1 3 ⇒ − = − (kL)1 sin(kL)1 tan(kL)1 I2 2 If
I1 /I2 = 1
EI L2 then v1 = v2
⇒ kL ≈ 4.057 ⇒ Pcr = 16.5 4. If there is no spa e between the bolt and the sleeve,
⇒
(
M1 = −EI1 v1′′ = P v1 − M0
M2 = −EI2 v2′′ = −P v2 + M0
⇒ EI1 v1′′ + EI2 v2′′ = 0 ⇒ v1′′ = 0
⇒ v1 = Ax + B
⇒ v1 ≡ 0 (since v1 (0) = v1 (L) = 0)
Bu kling is not possible. The situation is line in a pretensioned on rete beams.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
43
Example 2.10
What is the bu klng load of a beam with rounded ends. The length to height
ratio is L/h = 20.
P @ @ b b
EI
?h 6
L Solution:
EIv + P v = 0 · vˆ ZL ZL ⇒ EI v (4) vˆdx + P v ′′ vˆdx = 0 (4)
′′
0
⇒ −EI
0
ZL 0
v (3) vˆ′ dx − P
ZL
v ′ vˆdx = 0
0
(M = −EIv ′′ ) L ZL ZL ′ ′′ ′′ ⇒ M vˆ + EI v vˆ dx − P v ′ vˆ′ dx = 0 0
0
0
⇒ M(L)ˆ v ′ (L) − M(0)ˆ v ′ (0) + EI
ZL
v ′′ vˆ′′ dx − P
0
ZL
v ′ vˆ′ dx = 0
0
Boundary onditions:
M(0) + P Rv ′(0) = 0 and orrespondingly, when
x=L
M(L) − P Rv ′(L) = 0
Let's hoose
v(x) = v0 sin πx , vˆ(x) = sin πx L L ′
′
′
′
⇒ P R[v (L)ˆ v (L) + v (0)ˆ v (0)] + EI
ZL
′′ ′′
v vˆ dx − P
ZL
v ′ vˆ′ dx = 0
0 0 π 2 L π 2 π 4 L − P v0 + 2P Rv0 =0 ⇒ EIv0 L 2 L L 2 π 2 L EI R ⇒ v0 π2 2 − P 1 − 4 =0 L 2 L L π 2 EI ⇒ P = L2 1 − 4R L L 10 EI if R = ⇒ P = π2 2 40 9 L
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
44
Example 2.11
Determine the riti al load of the given beam by the nite element method.
Use two Euler-Bernoulli beam elements.
2EI
EI
L/2 Solution: dom:
@b b @
L/2
The olumn is divided into two elements, thus the model has three degrees of free-
v2 , ϕ2 , ϕ3 . 1j
Denoting where
K
P = λEI/L2 .
2j
2 ? 1
L/2
λSx,
P
3 ?P @ @ b b
L/2
We get a generalized linear algebrai eigenvalue problem
is the linear stiness matrix
λS
Kx =
is the gemetri stiness matrix, or initial stress
matrix. The element matri es of the Euler-Bernoulli model are
K
λS
(e)
(e)
EI = L
(e) ˜ =N
12 L2
6 L
4
− L122 − L6 12 L2
symm. 6 5L
1 10 2L 15
6 − 5L 1 − 10 6 5L
symm.
and the lo al degrees of freedom are thus in the order
6 L
2 6 −L 4 1 10 L − 30 1 − 10 2L 15
v1 , v1′ , v2 , v2′ .
Now
˜ (1) = N ˜ (2) = P = N
λEI/L2 . The onne tion from lo al to lo al degrees of freedom is:
(2) ϕ1 , δ3
=
(1)
δ1 = v2
(2)
(1)
= v1 , δ2 = ϕ2
=
(2) ϕ2 . The elements in the global matri es are:
8 · 12 EI 8 · 12 2EI + EI = 288 3 3 3 L L L EI (1) (2) = K34 + K12 = −24 2 L EI (2) = K14 = 24 2 L EI (1) (2) = K44 + K22 = 24 L (1)
(2)
K11 = K33 + K11 = K12 K13 K22
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
45
(2)
EI L EI =8 L
K23 = K24 = 4 (2)
K33 = K44
(1)
(2)
(1)
(2)
S11 = S33 + S11 = P
24 24EI =λ 5L 5L3
S12 = S34 + S12 = 0 EI 1 (2) =λ S13 = S14 = P 10 10L2 2EI 2L (1) (2) =λ S22 = S44 + S22 = P 15 15L L EI (2) S23 = S24 = −P = −λ 60 60L EI L (2) =λ S33 = S44 = P 15 15L Written in a matrix format:
288 −24 24
EI −24 L 24
where it is repla ed
24 4
4 8
δ1 /L δ2 δ3
24 5
= λ EI 0 L
1 10
0 2 15 1 − 60
1 10 1 − 60 1 15
δ1 /L
δ2 δ3
,
δ1 → δ1 /L and the uppermost equation is divided by L. From this eigenvalue
λi 's and the orresponding eigenve tors 1 24 −24 24 − λ 10 288 − λ 5 2 1 = 0 ⇒ λ det −24 24 − λ 15 4 + λ 60 kr = 26.32 1 1 1 4 + λ 60 8 − λ 15 24 − λ 10
problem we an solve
Let's investigate the onvergen e of the numeri al solution and how we an esitimate the error in our nite element solutions and how we an extrapolate an estimate of the exa t solution. If we know a priori the asymptoti onvergen e rate of the desired quantitity and the element in question, we an have have an improved estimate by omputing the problem at least by two dierent dis retizations, i,e meshes. The error in the numeri al solution in proportional to
h
Chk , where C
is a positive onstant,
is the mesh parameter (i.e. the hara teristi length of the largest element) and
k
is the rate
of onvergen e of the quantity in question. As an example, let's onsider the estimation of the exa t value of the load parameter by using extrapolation. The nite element solutions
λ1 = λ(h1 )
and
λ2 = λ(h2 )
satify
λ1 = λex + Chk1 λ2 = λex + Chk2 Rak-54.131 Stability of stru tures exer ises / 12.12.2005
46
First we eliminate equation, gives
C = (λ1 − λex )h−k 1
from the upper equation, and substituting it to the lower
h2 λ2 = λex + (λ1 − λex ) h1 from where we an solve
λex
k
k
λ2 − λ1 hh12 = k 1 − hh21
If we solve the same problem by using 10 elements (both parts of the beam have 5 equal elements), we get the value
λ = 25.18
for the ritial load. The rate of onvergen e of the
eigenvalues of the Euler-Bernoulli beam is
k=4
From this data we get the extrapolated value
and now
h1 = L/2, h2 = L/10 ⇒ h2 /h1 = 0.2.
λ1ex = 25.18,
whi h is also obtained by using 50
elements for the beam.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
47
Example 2.12
Determine the bu kling load of the stru ture below by the nite element
method and using one element for a member. The members an be assumed axially in nitely sti i.e. EA = ∞. P
P
?
? EI
EI
EI
e
Solution:
L
2L
e
Let's rst hoose the global o-ordinate system.
EA = ∞ ⇒ u2 = u3 = 0 6u ϕ2 2
?
v2
2j
6u3 ? ϕ3 3j
v3
v2 = v3
By symmetry
ϕ1 = ϕ4 ϕ2 = ϕ3
ϕ1 ?e
1j
ϕ4 ?e
4j
Therefore
the
bu kling
mode is antisymmetri
The global degrees of freedom are
ϕ1 , v2 , ϕ2 .
The elements of the global stiness matrix
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
48
are
(1)
(3)
(1)
(3)
(1)
(3)
(1)
(3)
(1)
(3)
(1)
(2)
EI 4EI =4 2L L 6EI EI = −2 = −3 2 2 (2L) L EI 2EI =2 =2 2L L 12EI EI =2 =3 3 3 (2L) L 6EI EI = −2 = −3 2 2 (2L) L
K11 = K22 + K22 = 2 K12 = K23 + K23 K13 = K24 + K24
K22 = K33 + K33 K23 = K34 + K34
(2)
(2)
(2)
(3)
K33 = K44 + K22 + K44 + K24 + K42 + K44 = 2
Note that the term
K33
4EI 2EI EI 4EI +2 +2 = 16 2L L L L
has the mixed terms 24 and 42 of the element 2.
The elements of the global geometri stiness matrix are
˜e Sije = N
The last element
S33
RL 0
Ni′ Nj′ dx
(1)
(3)
(1)
(3)
(1)
(3)
(1)
(3)
(1)
(3)
(1)
(3)
S11
= S22 + S22
S12
= S23 + S23
S13
= S24 + S24
S22
= S33 + S33
S23
= S34 + S34
S33
= S44 + S44
2 · 2L 8 EI P = λ 15 15 L 1 EI 1 = −2 P = − λ 2 10 5 L 2L 2 EI = −2 P = − λ 30 15 L 6 EI 6 P = λ 3 =2 5 · 2L 5 L 1 1 EI = −2 P = − λ 2 10 5 L 2 · 2L 8 EI =2 P = λ 15 15 L =2
does not have any terms from the element 2, sin e the normal for e of
that parti ular element is zero. The generalized linear algebrai eigenvalue problem is thus
4
− L3
2
ϕ1
EI 3 3 = λ EI −3 v − 2 2 L L L L L 3 ϕ2 2 − L 16
8 15 1 − 5L 2 − 15
1 − 5L
6 5L2 1 − 5L
2 − 15
ϕ1
1 v − 5L 2 8 ϕ2 15
These matri es an be non-dimensionalized by using a dimensionless displa ement and multiplying the se ond equation by
L.
δ2 = v2 /L
The lowest eigenvalue, i.e. the bu kling load fa tor
is
λkr = 0.528 ⇒ Pkr = 0.528 Rak-54.131 Stability of stru tures exer ises / 12.12.2005
EI L2 49
Example 2.13
Determine the riti al load of the stru ture shown below using the nite ele-
ment method and dividing the beam into two Euler-Bernoulli beam elements. Both ends of the beam are fully lamped. Determine as a fun tion of β , espe ially the ase α = 2 and 1 ≤ β ≤ 4. Hint:
Determine rst the distribution of the axial for e.
α β
EI
EI
EA
EA
L/2
Solution:
@ @ @ @ @ @ @ @
P
L/2
The normal for e distribution by solving the dierential equation of the axial dis-
pla ement:
(EA)i u′′ = 0 ⇒ u 1 = C1 x + C2
u 2 = C3 x + C4
Assuming linear elasti material, we get the for e-de e tion relationship:
Ni = (EA)i u′ =
(
βEAC1 − L2 ≤ x < 0
EAC3
0 0, the spring will reload elasti ally, i.e. Ei = E . 2
E1 = E, E2 = ET ⇒ P1 = PR = 2ER aL ,
where
ER = 2(EET )/(E + ET ).
In the ase of plasti bu kling the se ondary paths are stable when when
is
P0 ∈ (PT , PE ).
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
P0 ∈ (PT , PR )
and unstable
60
Example 3.4
In the beam theory taking the average transverse shear deformations into a -
ound, the equilibrium equations are
−EIθ′′ − kGA(v ′ − θ) = 0 kGA(v ′ − θ)′ − P v ′′ = 0
(49)
Show that in the ase of entrally ompressed antilever olumn the riti al load is
Pcr =
PE , 1 + αPE
where PE is the Euler bu kling load PE = π 2 EI/(4L2 ) and α = 1/(kGA) and k = 1/ζ .
P
Solution :
Equations (49) are the equilibrium equations
(
M′ − Q = 0
Q′ − P v ′′ = 0
expressed in terms of the kinemati al quantites
v and θ. The bending moment and the transverse
shear for e an be expressed as
( if
EI, kGA
and
P
M = −EIθ′
Q = kGA(v ′ − θ)
⇒ (49)
are onstants.
Dividing the equations (49) by
kGA
we get
v ′ − θ + αEIθ′ = 0
(1 − αP )v ′′ − θ′ = 0 Boundary onditions:
v(0) = 0
``` Q H H H QQ P
Q(0) = 0 M(L) = −EIθ(L) = 0
v′
Q(L) = P v ′ (L)
Let's try the solution in the form
⇒
v = Aerx , θ = Berx [Ar + B(αEIr 2 − 1)] erx = 0
[(1 − αP )Ar 2 − Br] erx = 0
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
61
⇒
"
r (1 − αP )r 2
αEIr 2 − 1 −r
To have a meaningfull nontrivial solution (A
#
A B
6= 0 6= B ),
!
0
=
0
!
it is required
det = 0 ⇒ −r 2 1 + (1 − αP )(αEIr 2 − 1) = 0
⇒ r1,2 = 0 tai 1 + (1 − αP )(αEIr 2 − 1) = 0 −P ⇒ r2 = (1 − αP )EI
Now
P > 0,
what about
(1 − αP ) = 1 − P/kGA?
Denoting
P = λEI/L2
E I EI = 2(1 + ν)k −1 (inserting G = , k −1 = η) 2 2 kGAL AL 2(1 + ν) I 1 2(1 + ν)η , 0 ≤ ν ≤ , η ≈ 1, I/A GIt r −2 .
P r 2 − GIt EIω
The general solution of this equation is
ϕ = A + Bx + C cos kx + D sin kx and the boundary onditions are
ϕ(0) = 0 ⇒ A + C = 0
ϕ(L) = 0 ⇒ A + BL + C cos kL + D sin kL = 0
ϕ′′ (0) = 0 ⇒ C = 0 ⇒ A = 0
ϕ′′ (L) = 0 ⇒ C cos kL + D sin kL = 0 ⇒ D sin kL = 0 ⇒ B = 0 Sin e
A = B = C = 0,
we get
nπ L P r 2 − GIt nπ 2 = ⇒ EIω L 2 GIt + EIω Lπ 2 ⇒ Pcr,ϕ = r2
sin kL = 0 ⇒ k =
Let's insert the inertias
It
bt 2 (b + t2 ) 12 = 0.78bt3
Iω
= 0
Iz
= Iy =
⇒ Pcr,ϕ = Now when
Iω = 0
It t2 GA GA = 4.67 2 GA = 4.67 2 Iz + Iy b +t 1 + (b/t)2
the torsional b kling load is onstant and does not depend on the length of
the beam. The torsional bu kling load for the ase in question ould be obtained dire tly from equation (57)3 by setting
Iω = 0 (P r 2 − GIt )ϕ′′ = 0 ⇒ Pcr,ϕ = GIt /r 2
What is the bu kling mode in this ase?
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
89
Example 4.2
Determine the riti al load Pcr for a entrally ompressed lamped beam. The
ross-se tion is shown below and b = 10t, ν = 0. Determine the riti al load as a fun tion of the length. ? t 6
b
b
Solution:
The dierential equations for torsional bu kling for a olumn are
EIz v (4) + P (v ′′ + zv ϕ′′ ) = 0
EIy w (4) + P (w ′′ − yv ϕ′′ ) = 0 EI ϕ(4) − GI ϕ′′ + P (z v ′′ − y w ′′ + r 2 ϕ′′ ) = 0 ω t v v For a T-beam we have
-
b
d= 6 ? -
6
z
b ?
y
?
b 4
zv = 0,
EIω = 0
yv = −b/4 5 b3 t , Iz = b3 t Iy ≈ 12 24 Ip 5 2 3 + yv2 + zv2 = b2 It ≈ t b, r 2 = 3 A 24
The equations simplify now to the form
EIz v (4) + P v ′′ = 0 EIy w (4) + P [w ′′ − yv ϕ′′ ] = 0
−GIt ϕ′′ + P [−yv w ′′ + r 2 ϕ′′ ] = 0
The upper equation, i.e. the displa ement in the
z -dire tion
y -dire tion
un ouples from the displa ement in
and from the twist-rotation, thus the bu kling in
Py = 4π 2
y−dire tion
gives the load
EIz L2
Fun tion whi h satisfy the boundary onditions are
nπx w = B 1 − cos 2 L nπx ϕ = C 1 − cos 2 L
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
90
Let's denote
P = λGIt r −2 GIt r2
where
"
α = 4π 2 (r/L)2 EIy /GIt .
α−λ
yv λ r 2 (1 − λ)
yv λ
#
B C
!
!
0
=
0
In order to have a non-trivial solution for
has to vanish. The riti al value for the
λ
A, B
the determinant
parameter an be found by solving the hara teristi
polynomial
If we denote
(yv /r)2 =
Iy = I ,
then
y 2 v λ2 − (1 + α)λ + α = 0 1− r
Iz = 52 I
and
2 I . If 25
ν=0
then
G = E/2
and
GIt =
1 EI . Also 25
3 , thus the hara teristi polynomial has the form 10
λ2 − where
It =
α=
10 10 (1 + α)λ + α = 0 7 7
125 2 π (b/L)2 . The smaller toot is 6
λ1 = 57 (1 + α) 1 − Note, that
λ1 ≤ 1 . Py =
s
14α 1− 5(1 + α)2
!
The bu kling load is now the minimum from
EIz 4π 2 2 L
2 r 2 GI GIt 625 2 b t = 250π = π 2 L r 12 L r2 2
Pz,φ,1 = λ1
z−dire tion 2 GIt GIt 125 2 b 2 EIy π = Py > Pz,φ,1 Pz = 4π 2 2 = α 2 = L r 6 L r2 5
Note that, if the torsional mode is prevented the bu kling load in
The riti al load parameter
λcr = λ1
GIt r2
is
is shown below as a fun tion of the slenderness
(L/b)
1 0.9 0.8 0.7 0.6
λcr 0.5 0.4 0.3 0.2 0.1 0
0
20
40
60
80
100
L/b Rak-54.131 Stability of stru tures exer ises / 12.12.2005
91
What is the riti al load for a olumn lamped at its lower end and the upper
Example 4.3
end free. The ompressive for e a ts on the enter or gravity. The hight of the olumn is 1000 mm and the material's Young's modulus is E = 210 GPa, the Poisson ratio ν = 0.3 and the yield strength σm = 220 MPa. The ross se tion is shown below and the warping an take pla e freely at the upper end of the olumn. The hight of the ross-se tion is h = 100 mm and the wall thi kness is t = 10 mm. ? t 6
h
h Solution:
The orresponding dierential equation system is
EIz v (4) + P [v ′′ + zv )ϕ′′ ] = 0 EIy w (4) + P [w ′′ − yv )ϕ′′ ] = 0
EIz ϕ(4) − GIt ϕ′′ + P (zv v ′′ − yv w ′′ + r 2 )ϕ′′ ) = 0
where
(yv , zv )
are the oordinates of the shear enter and
r 2 = yv2 + zv2 + (Iz + Iy )/A.
For the
C-se tion pro le in question we have
• Iy = 13 h3 t • Iz =
7 3 ht 12 3
• It = ht
• A = 3ht
We have to ompute the shear enter oordinates, therefore we need the se torial quantities.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
92
Let's hoose the pole B as the orner point
ωB = ωB (s) =
s = 2h
s =hhhh
hhh −h2 hhh hh h
Zs
h(s)ds
0
−h ≤ s ≤ h , ωB = 0
h ≤ s ≤ 2h , ωB = h(h − s)
When omputing the se torial oordinate
ω
the sign is
positive when we rotate ounter lo kwise and negative
s=0 s=h
B
when we are moving in a lo kwise dire tion.
The shear enter oordinates are
yv = yB + Izω /Iy = 0 (symmetry) zv = zB + Iyω /Iz
Next we determine
Iyω
−h/2
−h/2
Iyω y
h/2 Now the se torial oordinate
×
ωv′
` 3 2 ```` h ``` 7 ``` ` DD D D D V D ωv′ D D 3 D h 7 D D D B ```` ``` ``` ` `
Adding a onstant
C
s
th4 h = y(s)ωB (s)t(s)ds = t − h(h − s)ds = 2 4 h h 16 1 47 3 ⇒ zv = − − ht h t =− h 3 4 12 21 521 I + I z y = h2 ⇒ r 2 = yv2 + zv2 + A 588
wrt the shear enter.
1 2 h − 14
The
ωv -diagram
above is not nal, sin e it has to ful ll
the normalizing ondition
I
1 2 h 2
ωv (s)t(s)ds = 0
s
ωv -diagram, we have I I I ′ ′ ωv tds + Ctds = ωv tds + C tds = 0
to the
I
Z2h
Z
s
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
s
93
⇒C = −
If we hoose the origin of the
3 2 h 14 ````
DD D
×V
``` ``` ` `
D D
D
s- oordinate
6
3 − 14 h
ωv′ tds 3 = − h2 A 14
the point
y = 0, z = −h/3.
h 3 h − 28 (14s + 13h) , − 2 h ≤ s ≤ − 2 3 ωv (s) = hs , − h2 ≤ s ≤ h2 3 − h (14s − 13h) , h ≤ s ≤ 3 h 28 2 2
ωv
D D `D `` ``` D ``` `` 2 `
s
− 27 h2
s
D D
H
2 2 h 7
The warping rigidity is
Iω =
Z
A
ωv2dA = 2t
Zh/2 0
9 2 2 h s ds + 49
3h/2 Z
h/2
1 2 5h5 t h (14s − 13h) = 282 84
Now we an investigate the stability of the olumn
P
?
The boundary onditions are
L
v(0) = w(0) = ϕ(0) = 0
x 6
v ′ (0) = w ′ (0) = ϕ′ (0) = 0 v ′′ (L) = w ′′ (L) = ϕ′′ (L) = 0
Fun tions satisfying the above boundary onditions are
πx πx πx − 1 , w = C2 cos − 1 , ϕ = C3 cos −1 v = C1 cos 2L 2L 2L Inserting these into the dierential equations we get
π2 EIz − P C1 − P zv C3 = 0 2 4L2 π EIy − P C2 = 0 4L2 2 π 2 EIω + GIt − P r C3 = 0 −P zv C1 + 4L2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
94
Let's denote
π2 π2 π2 Py = EI , P = EI , P = EIω + GIt z z y ϕ 4L2 4L2 4L2 0 C1 Py − P 0 −zv P ⇒ 0 Pz − P 0 C2 = 0 0 C3 −zv P 0 Pϕ − P r 2
In order to have a nontrivial solution we must have
det[·]=0
⇒ (Pz − P ) (Py − P )(Pϕ − P r 2 ) − zv2 P 2 = 0 ⇒ P1 = Pz
tai (r 2 − zv2 )P 2 − (Pϕ p + Py r 2 )P + Py Pϕ = 0 (Pϕ + Py r 2 ) ± (Pϕ + Py r 2 )2 − 4Py Pϕ (r 2 − zv2 ) ⇒ P2,3 = 2(r 2 − zv2 )
Inserting the values we get the gures
P1 = 1727 kN, P2 = 964 kN, P3 = 11454 kN ⇒ Pkr = P2 Pkr σ= = 321 MPa > σm = 220 MPa A ⇒ Pkr ≤ 660 kN So, the yielding takes pla e prior to bu kling, therefore we should determine the plasti -bu kling load.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
95
Example 4.4
A beam with a re tangular ross-se tion is loaded by a onstant bending moment
M . The lower boundary of the ross-se tion annot move in lateral dire tion but an rotate. The rotation is supressed at the simply supported boundaries. Determine the riti al bu kling moment Mcr .
M
M
b b 6666666666666666666666@ @ @ @ b b
L = 50b
Solution:
b b
b
In this ase the dierential equations are
(
EIy w (4) − Mz0 ϕ′′ = 0
−GIt ϕ′′ − Mz0 w ′′ = 0
Dierentiating the lower equation twi e and multiplying by parts the rst equation by
(58)
Mz0 /EIy
we get
2
(M 0 ) − z ϕ′′ + Mz0 w (4) = 0 EIy −GIt ϕ(4) − Mz0 w (4) = 0 Adding these two equations, gives
2
(M 0 ) − z ϕ′′ − GIt ϕ(4) = 0 trial ϕ = erx EIy 0 2 (M ) ⇒ − z r 2 erx − GIt r 4 erx = 0 EIy 2 (Mz0 ) 2 ⇒r =− EIy GIt Denoting
k 2 = −r 2
we get
ϕ = A1 sin kx + A2 cos kx + A3 x + A4
(59)
w = B1 sin kx + B2 cos kx + B3 x + B4
(60)
In a similar fashion we get
Substituting equations (59) and (60) to equations (58) we get
B1 = −A1
GIt GIt , B = −A 2 2 Mz0 Mz0
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
96
Sin e the lower boundary is immovable in
w+ Hen e
z -axis
dire tion
5b w 5b ϕ≡0 ⇒ =− 2 ϕ 2
(61)
w B2 5b B1 2GIt = = − ⇒ Mz0 = ≡ ϕ A1 A2 2 5b
The bu kling mode requires
M0 nπ n halfwaves k=p z = L EIy GIt 2GIt nπ p EIy GIt = L 5b
⇒ Mz0 =
Let's insert the ross-se tional values
Iy = It = L = ⇒ Mz0 = ⇒n =
5 5b · b3 = b4 12 12 5 1 3 5b · b = b4 3 3 50b and if G = 0.4E nπ 4 p Eb 0.5 · 5/12 · 5/3 = 0.0331nEb3 = 0.267Eb3 50b 8.
The riti al moment an be obtained by substituting equation (61) into equation (582 ):
′′
⇒ −GIt ϕ − Now
ϕ′′ 6= 0,
sin e if
A=B=0⇒ϕ≡0
ϕ′′ = 0 ⇒ ϕ = Ax + B .
Mz0
5b − ϕ′′ = 0 2
Taking boundary onditions into a
ount, we get
and bu kling is not possible hen e
−GIt −
Mz0
5b − 2
= 0 ⇒ Mz0 =
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
2GIt 5b
97
Example 4.5
Determine the riti al lateral bu kling moment Mcr for the beam shown below.
The support on the rhs side prevents verti al and lateral displa ements but the ross-se tion
an rotate about the support. The ross-se tion is re tangular with dimensions b × h where
h ≫ b.
M
M
b @ @
Solution:
b @ @ b b
The dierential equations takes now the form
(
EIy w (4) − Mz0 ϕ′′ = 0
(62)
−GIt ϕ′′ − Mz0 w ′′ = 0
Boundary onditions on the lhs support
w(0) = 0, w ′′ (0) = 0, ϕ(0) = 0 The rhs boundary onditions are slightly more ompli ated
ϕ A ? A A A M * z¯ -z, w A A A A AU A Ay ¯A A @ @
The kinemati al onstraint at the enter of gravity of the ross-se tion is
?y, v
h w(L) = − ϕ(L) 2
A M z¯ K K My¯-z A M y ?
Let's divide the external moment
M
into
omponets parallel to the deformed oordinate axis
6z
Mx¯ w ′ (L) : C M CCW z¯ ? M
-
z
Mz¯ ≈ M
My¯ = EIy w ′′ ≈ −ϕ(L)M
Mx¯ = −w ′ (L)M
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
98
The boundary onditions are
w(0) = 0 w ′′ (0) = 0 ϕ(0) = 0
h w(L) = − ϕ(L) 2 −EIy w ′′ (L) = −ϕ(L)M GIt ϕ′ (L) = −w ′ (L)M
Substituting equation (622 ) into equation (621 ) saadaan
w
(4)
M2 + k w = 0, k = EIy GIt ⇒ w = A sin kx + B cos kx + Cx + D 2
2
′′
From boundary onditions we get
w(0) = w ′′ (0) = 0 ⇒ D = B = 0 ⇒ w = A sin kx + Cx
From the dierential equation (622 ) we an dedu e that
ϕ
is of similar form
⇒ ϕ = E sin kx + F x
⇒ −GIt k 2 E sin kx − Mk 2 A sin kx = 0 M ⇒ E = − GI A t
Let's substitute the boundary onditions into these trial fun tions
M A cos kx − F ) = −(Ak cos kx + C)M GIt M ⇒F =− C GIt h M h (A sin kL + CL) w(L) = − ϕ(L) ⇒ A sin kL + CL = 2 2 GIt Mh Mh ⇒ 1− A sin kL + 1 − CL = 0 2GIt 2GIt M (A sin kL + CL) −EIy w ′′ (L) = −ϕ(L)M ⇒ EIy k 2 A sin kL = GI t M2 M 2 ⇒ EIy k − A sin kL + − C=0 GIt GIt GIt ϕ′ (L) = −w ′ (L)M ⇒ −GIt (k
Sin e
k 2 = M 2 /EIy GIt
it follows from equation (63)
(63) we obtain
Mh 1− 2GIt
−(M/GIt )C = 0 ⇒ C = 0. From equation
sin kL = 0
The riti al moment is then
Mcr = min
(
2GIt , π h
p
EIy GIt L
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
) 99
The eigenmodes are
w(x) = A sin kx M ϕ(x) = − A sin kx GIt Note! if
Mcr =
2GIt h
⇒ kL 6= π ⇒ w(L), ϕ(L) 6= 0.
If
Mcr =
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
2GIt h
⇒ kL = π .
100
Example 4.6
Determine the riti al bu kling moment in the form Mcr = λ
p EIy GIt /L, where
the parameter λ = λ(k, h/L). Draw the riti al load parameter λ as a fun tion of k , when
k ∈ (−1, 1) and L/h = 20, ν = 0. Use the prin iple of minimum potential energy or some other numeri al method and use trigonometri trial fun tions.
M
kM
b @ @
b @ @ b b
L Solution by the prin iple of minimum potential energy:
h
b
The expression for the total potential
energy is
1 Π = 2
ZL 0
GIt (ϕ′ )2 + EIy (w ′′ )2 + 2(Mz0 ϕ)′ w ′ dx
ZL i 1 h ′ GIt (ϕ′ )2 + EIy (w ′′ )2 + 2(Mz0 ϕ + ϕ′ Mz0 )w ′ dx = 2 0
where
′
Mz0 = M(1 − x/L) + kMx/L = M[1 + (k − 1)x/L] ⇒ Mz0 = (k − 1)M/L.
trial fun tions
ZL
ϕ = ϕ0 sin πx/L, w = w0 sin πx/L.
Let's use the
Hen e
′
Mz0 ϕw ′dx = 0
0
ZL
x πx L cos2 dx = L L 4
0
π 2 L π 4 L π 2 L L 1 2 2 GIt ϕ0 + EIy w0 + 2M ϕ0 w 0 + (k − 1) ⇒Π = 2 L 2 L 2 L 2 4 Minimizing the potential energy
∂Π ∂ϕ0 ∂Π ∂w0
π 2 L π 2 GIt L = =0 ϕ0 + M + (k − 1) w0 2L L 2 4 π 2 π 4 EIy L L =0 = w + M ϕ + (k − 1) 0 0 3 2L L 2 4 ! ! " # π 2 EIy M k−1 w 0 1 + 0 3 2L = ⇒ 2L GI 2 k−1 M t ϕ0 0 1 + 2L 2 2L
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
101
The riti al moment is obtained in the form
π Mcr = ± 1 1 + 2 (k − 1) When
k = 1
p
we get the exa t result. When
GIt EIy = λ(k) L
p
GIt EIy L
k → −1 ⇒ λ → ∞,
thus the trial fun tion is
inadequate to model su h a situation. How the trial fun tion should be sele ted to result in a
4π
λ
3π
2π
π
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
k meaningful solution for the ase
k = −1?
Solution by the method of weighted residuals:
Let's start from the dierential equations
EIy w (4) − (Mz0 ϕ)′′ = 0
−GIt ϕ′′ − Mz0 w ′′ = 0.
Multiplying the rst equation by the lateral displa ement weight fun tion
φˆ and integrate over the domain Z L (4) w[EI ˆ − (Mz0 ϕ)′′ ]dx = 0 yw 0 Z L ′′ 0 ′′ ϕ(−GI ˆ t ϕ − Mz w )dx = 0.
by the rotation weight fun tion
wˆ
and the lower one
we get
0
Integrating by parts
L L Z L ′′ 0 ′ ′′′ 0 ′ [EIy w − (M ϕ) ]wˆ − (EIy w − M ϕ)wˆ + (w ˆ ′′EIy w ′′ − wˆ ′′ Mz0 ϕ)dx = 0 z z 0 0 0 L Z L ′ (ϕˆ′ GIt ϕ′ − ϕM ˆ z0 w ′′)dx = 0. − ϕˆ GIt ϕ + 0
0
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
102
The boundary terms wanish and if we use same basis for the weight as for the trial ones
ψ(x) = sin(πx/L): w = w0 ψ(x),
ϕ = ϕ0 ψ(x),
wˆ = wˆ0 ψ(x),
and
ϕˆ = ϕˆ0 ψ(x),
we get the equation
wˆ0 ϕˆ0
!T "
K11
MG12
MG21
K22
#
w0 ϕ0
!
=0
where
G12
K22
L
L π 4 EIy π4 EIy (ψ (x)) dx = EIy 4 sin2 (πx/L)dx = L 2L3 0 0 Z L Z L π2 = G21 = − [1 + (k − 1)x/L] ψ(x)ψ ′′ dx = [1 + (k − 1)x/L] 2 sin2 (πx/L)dx L 0 0 2 k−1 π 1+ = 2L 2 Z L π 2 GIt = GIt (ψ ′ (x))2 dx = 2L 0
K11 =
Z
′′
Z
2
So we got the same stiness matrix as in the potential energy approa h. Let's try to write the stiness matrix in a dimensionless form. First we write it in the form
wˆ0 /L ϕˆ0 where
Denoting
!T "
˜ 11 M G ˜ 12 K ˜ 21 K ˜ 22 MG
#
w0 /L ϕ0
!
=0
2 k−1 π2 π 4 EIy ˜ 22 = K22 π GIt ˜ ˜ 1+ , K , G12 = K11 = 2L 2 2 2L p M = λ EIy GIt /L and η 2 = EIy /GIt we obtain the matrix in a dimensionless wˆ0 /L ϕˆ0
!T
π2
p
EIy GIt 2L
"
π2η
λ 1+
The riti al value for the load parameter
λ
k−1 2 −1
λ 1+
k−1 2
η
#
w0 /L ϕ0
!
form
=0
is then
λcr = ±
π 1+
1 (k 2
− 1)
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
103
Example 4.7
Determine the riti al moment Mcr for the beam shown below, the proportions
are b = 10t, L = 20b, ν = 1/3. What is the result if M is negative?
M
M
b @ @
b @ @ b b
L Solution:
b @ I @ @b
@ R @ @ @
The ross-se tional onstants are
√ 1 3 1 3 2 2 3 b, zv = 0 It = t b, Iy = tb , Iz = tb , yv = − 3 Z 3 12 Z 4 Z √ √ √ bt 2 1 2 1 2 4 2 2 3 2 y(y + z )dA − 2yv , y dA = 0, yz dA = 2 b b = tb ⇒ βz = 2b βz = Iz 6 4 2 24
The dierential equations for the lateral/torsional bu kling are
EIy w (4) − Mϕ′′ = 0
−GIt ϕ′′ − Mw ′′ − βz Mϕ′′ = 0 w (4) +
⇒
ϕ′′ = −
M w ′′ GIt + βz M
M2 w ′′ = 0 EIy (GIt + βz M)
The general solution is
w = A sin kx + B cos kx + Cx + D
k2 =
where
M2 EIy (GIt + βz M)
The boundary onditions are
w(0) = 0 ⇒ B + D = 0
w ′′ (0) = 0
B=0
w(L) = 0
A sin kL + CL = 0 Ak 2 sin kL = 0
′′
w (L) = 0
The lowest bu kling load is obtained when
M 2 − βz denoting
p M = λ EIy GIt /L
and
n = 1,
⇒
kL = nπ,
hen e
π2 π2 EI M − EI GI y y t 2 = 0 L2 L
EIy = α2 GIt λ2 − π 2 α
βz − λ − π2 = 0 L
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
104
The roots are
2
λ= Substituting
βz =
√
π α 2
βz L
1 ±
s
2 b, L = 20 b, α2 = 4000/3, λ = 2.62π 2
1+
4 π 2 α2
L βz
gives the result
∨
2
λ = −0.04π 2
k 2 is positive for negative λ values, i.e. if it holds GIt +βz M > 0. p √ ! EIy GIt 5 = GIt 1 + 2 √ λ = −0.02 GIt + βz λ L 3
Let's he k if the expression for
Therefore the trial fun tion for
w
is wrong for a negative moment. In this ase
M2 EIy (GIt + βz M) w(x) = A sinh kx + B cosh kx + Cx + D
w ′′′′ − k 2 w ′′ = 0, where k 2 = −
From boundary onditions we get
B=D=0
A sinh kL + CL = 0 Ak 2 sinh kL = 0 Sin e
k 6= 0
and
!
⇒A=C =0∨k =0
the beam does not bu kle laterally. However, the anges an bu kle in a plate-like
mode.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
105
5
Bu kling of plates
Example 5.1
A square plate is stiened by equidistant beams of re tangular ross-se tion in
the loading dire tion. How many stieners are required to obtain a bu kling load Nx at least 2
the value 10 πa2D . Thi kness of the plate is h, whi h is also the width of the beam. The height of the beams is αh = 4h. The material is isotropi with Poisson's ratio 0.3. Use the energy method and a one-parametri trial fun tion for the de e tion w(x, y). The plate is simply supported and the torsional stiness of the beams need not to be taken into a
ount. h = a/40, where a is the side-length of the plate.
Nx
-
a n+1
n kpl palkkeja
? y
Solution:
-
x
a
a
Let's use the following trial fun tion to the de e tion
w(x, y) = w0 sin
πy πx sin a a
Expression for the total potential energy of the plate is
∆Π = ∆UZ + ∆V = ∆Uplate + ∆Ubeams + ∆Vplate + ∆Vbeams D ∆Uplate = (∆w)2 dA 2 A
∆w = w,xx + w,yy , and w,xx D π4 2 w 2 a2 Z0 Nx Nx π 2 2 2 = − w w,x dA = − 2 2 4 0
πx πy π2 sin = w,yy = −w0 2 sin a a a
⇒ ∆Uplate = ∆Vplate
A
∆Ubeams =
n X i=1
∆Vbeams = −
EI 2
n X i=1
Za
2 w,xx dx =
0
σx hαh 2
Za
EI π 4 2 X 2 πi sin w 4 a3 0 n+1
2 w,x dx, where σx h = Nx
0
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
106
X Nx h πi = − α w02 π 2 sin2 n+1 4 4a Nx π 2 EI π 4 X 2 πi Nx h 2 X 2 πi Dπ w02 − + sin − α π sin ⇒ ∆Π = 2 3 2 a 2 4 4 a n+1 4 a n+1 When omputing the
∆Vpalkit
term, it is assumed that the load
Nx
is equally distributed for
the ross-se tional area of the beam. Using the notation
Nx = λ In the example ase
α=4
α3 h4 Eh3 π 2 Eh3 , I = , D = , when ν = 0 12a2 12 12
and
h = a/40.
" n X πi Eh3 π 4 2 3 h sin2 w0 1 + α −λ ⇒ ∆Π = 2 24 a 2a i=1 n+1 The equilibrium equations from the ondition
hara terized by
n
h X 2 πi 1 sin +α 4 2a i=1 n+1
δΠ = 0 ⇒ w0 = 0,
!#
and the riti al point is
∂2Π =0 ∂w02 P πi h sin2 n+1 1 + α3 2a P 2 πi ≥ 10 ⇒λ = 1 h sin n+1 + α 2a 4 δ2Π = 0 ⇒
Substituting
α=4
ja
h = a/40
n=1 ⇒ n=2 n=5 n=9
and trying dierent
πi n+1 X πi sin2 n+1 X πi sin2 n+1 X 2 πi sin n+1 X
sin2
n's:
1 + 54 1 1 = 6 + 20 4 1 + 45 32 3 3 =2· = ⇒λ= 1 1 3 ≈ 6.8 4 2 + 4 20 2 =1⇒λ=
= 3 ⇒ λ = 8.5 = 5 ⇒ λ = 10
Nine stieners will be su ient.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
107
Example 5.2
Determine τcr for an in nite plate strip using a trial fun tion
w(x, y) = A sin(πy/b) sin[π(x − αy)/s] where s is the half wavelength of the bu kling mode. The plate is simply supported and it's bending stiness is D . How large is the error in omparison to the analyti al solution τcr =
5.35π 2 D/b2 t (t is the thi kness of the plate)? τ
b - - - - - - - - - -
τ Solution:
Using the trial fun tion
w(x, y) = A sin where and
s
is the half wavelength in
x = αy + s
x-axis
πy π sin (x − αy) b s
dire tion. De e tion vanish (w
= 0)
at lines
x = αy
in addition to the boundaries.
s x
y αb The expression for the total potential energy is
D ∆Π = 2
Z
2
(∆w) dA + Nxy
A
Let's integrate a sli e between the lines
Z
w,x w,y dA
A
y = 0, y = b, x = αy
and
x = αy + s,
i.e. the area of
one half-wavelength:
π πy π w,x = A sin cos (x − αy) s b s Rak-54.131 Stability of stru tures exer ises / 12.12.2005
108
π2 πy π w,xx = −A 2 sin sin (x − αy) s b s πy π π πy π π sin (x − αy) − Aα sin cos (x − αy) w,y = A cos b b s s b s π2 πy π ππ πy π w,yy = −A 2 sin sin (x − αy) − Aα cos cos (x − αy) b b s bs b s 2 πy π π πy π ππ cos cos (x − αy) − Aα2 2 sin sin (x − αy) −Aα sb b s b s s 2 πy π π2 π2 2π sin (x − αy) ∆w = w,xx + w,yy = −A 2 + 2 + α 2 sin s b s b s ππ πy π −2Aα cos cos (x − αy) bs b s hπ i πy π πy π π πy π 2π w,x w,y = A sin cos (x − αy) cos sin (x − αy) − α sin cos (x − αy) s b s b b s s b s Change of variables
( Sin e det[
Zb 0
x = t + αr y=r
∂(x, y)/∂(t, r)
x−αy+s Z
Z b Zs 0
Z b Zs 0
0
"
xt yt xr yr
#
=
"
1 0 α 1
#
℄ = 1, the s ale is preserved.
(∆w)2 dxdy =
x−αy
∂(x, y) = ⇒ ∂(t, r)
(∆w)2 dtdr = A2
0
2
w,x w,y dtdr = −A
Z Z
α
"
π2 π2 π2 + + α b2 s2 s2
2
# 4 π bs + 4α2 2 (bs) 2 2
2 π2 2 πr 2 πt 2 π b sin cos dtdr = −A α s2 b s 4 s
# 2 2 4 π2 π2 b π bs π 2 2 − 2α (1 + α ) + 2 +α Nxy = 0 s2 b 4 bs 4 s s2 b2 π2D 2 2 2 2 + 6α + 2 + 2 (1 + α ) ⇒ Nxy = 2αb2 b s 2 2 s b π2D 2 2 2 2 + 6α + 2 + 2 (1 + α ) ⇒ τ= 2αb2 t b s
∂2 D ⇒ ∆Π = 2 ∂A2 2
"
The expression of the shear stress still ontains two free parameters obtained when
τ
α
and
s.
The minimum is
is minimized with respe t to these two paramaters:
π2D s2 b2 (1 + α2 )2 π2D 2 = + 6α + + f (α, s) τ = 2b2 t α b2 α s2 α 2b2 t ∂f 2s (1 + α2 )2 (−2b2 ) s √ = + = 0 ⇒ = 1 + α2 2 3 ∂s αb α s b 2 1 + α ⇒ f˜ = + 6α + 2 α α Rak-54.131 Stability of stru tures exer ises / 12.12.2005
109
∂ f˜ 2 2α2 − (1 + α2 ) 1 s √ = − 2 +6+2 = 0 ⇒ α = ± = ⇒ ∂α α α2 b 2 √ π2D π2D ⇒ τcr = 4 2 2 ≈ 5.66 2 bt bt
The dieren e to the analyti al value 5.35
r
3 2
π2 D , is thus 5.8 %. b2 t
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
110
Example 5.3
Derive the bu kling equations for a ir ular plate uniformly ompressed in the
radial dire tion.
Nr =-P
R
Solution:
Let's rst investigate the equilibrium equations of a plate loaded in its plane The
equilibrium equation in the radial dire tion is
The plate has
dϕ ∂Nr dr (r + dr)dϕ − Nr rdϕ − 2Nϕ dr = 0 Nr + ∂r 2 ∂Nr Nr − Nϕ + =0 ⇒ ∂r r 0 0 now a stress state Nr , Nϕ . Investigating the equilibrium in a slightly
state, gives
D
d4 w 2 d3 w 1 d2 w 1 dw + − − 3 4 3 2 2 dr r dr r dr r dr
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
de e ted
= p˜ 111
where
p˜ is
Nr0 , Nϕ0 : ∂ 0 dw 0 dw 0 dw rdϕ + Nr Nr dr (r + dr)dϕ + p˜rdrdϕ = −Nr dr dr ∂r dr ∂ ∂ 0 dw 0 dw 0 dw = Nr N rdrdϕ + Nr (dr)2 dϕ drdϕ + dr ∂r r dr ∂r dr N 0 dw d dw ⇒ p˜ = r Nr0 + r dr dr dr
the lateral omponent due to the membrane for es
The resulting dierential equation is thus
d4 w 2 d3 w 1 d2 w 1 dw 1 Nr0 dw d 0 dw Nr + − 2 2 − 3 = + dr 4 r dr 3 r dr r dr D r dr dr dr
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
112
Derive the potential energy expression for a radially ompressed ir ular plate:
Example 5.4
) # ′ 2 ZR ( " 2ν w + w ′w ′′ − P (w ′)2 rdr Π(w) = π D (w ′′ )2 + r r 0
Solution:
The strain energy for a plate is
U = U0
Z
uo dV = 2π V
ZR Zh/2
U0 dzrdr
0 −h/2
1 1 σr ¯ǫr + σϕ ǫ¯ϕ + σr0 ǫ¯r + σϕ0 ǫ¯ϕ = 2 2
where the strains have expressions
ǫ¯r = ǫr + zκr ǫ¯ϕ = ǫϕ + zκϕ
Zh/2 1 ⇒ U = 2π σr zdz + κϕ σϕ zdz 2 0 −h/2 −h/2 −h/2 −h/2 h/2 h/2 h/2 h/2 Z Z Z Z 0 0 0 0 dz rdr zdz + κϕ σϕ dz + κr σr dz + ǫϕ σϕ +ǫr σr ZR
= 2π
1 ǫr 2
Zh/2
1 σr dz + ǫϕ 2
Zh/2
−h/2
−h/2 R Z
1 σϕ dz + κr 2
−h/2
Zh/2
−h/2
1 1 1 1 0 0 ǫr Nr + ǫϕ Nϕ + κr Mr + κϕ Mϕ + ǫr Nr + ǫϕ Nϕ rdr 2 2 2 2
0
The potential of external loads is
V
= −2πRNr0 u(R) = −2πR Nr0 u(R) − Nr0 u(0) ZR d 0 0 = −2π Nr u(r) + [Nr u(r)]r dr dr 0
The bending moments for an isotropi plate are
Mr = D(κr + νκϕ ) Mϕ = D(κϕ + νκr )
⇒ ∆Π = π
ZR 0
κr D(κr + νκϕ ) + κϕ D(κϕ + νκr ) + 2(Nr0 ǫr + Nϕ0 ǫr ) rdr
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
113
+π
ZR
= πD
ZR
0
(ǫr Nr + ǫϕ Nϕ )rdr − 2π
ZR
Nr0 u
0
(κ2r + κ2ϕ + 2νκr κϕ )rdr + 2π
0
ZR 0
d 0 + (Nr u)r dr dr
ZR d 0 0 0 0 (Nr ǫr + Nϕ ǫϕ )rdr − 2π Nr u + (Nr u)r dr + π dr 0
Let's substitute the expressions
du 1 ǫr = + dr 2
dw dr
2
, ǫϕ =
d2 w 1 dw u , κr = − 2 , κϕ = − r dr r dr
and after some manipulations we get:
ZR
∆Π = πD
0
(w ′)2 2ν ′ ′′ (w ) + 2 + w w rdr r r ′′ 2
ZR 1 ′ 2 0 ′ 0 ′ +2π Nr u + (w ) − (Nr u) rdr 2 0
+2π
ZR
(Nϕ0
−
Nr0 )udr
+π
0
The expressions for
Nr0
and
Nϕ0
Nr =
ZR
(ǫr Nr + ǫϕ Nϕ )rdr
0
in terms of strains are
E (ǫ 1−ν 2 r
Nϕ =
+ νǫϕ), where
E (ǫ 1−ν 2 ϕ
+ νǫr )
du = u′ dr u ǫϕ = r
ǫr =
Taking the radial equilibrium equation into a
ount gives:
dNr Nr − Nϕ + =0 dr r u u 1 1 ′ ν u + ν − − νu′ = 0 ⇒ u′′ + u′ − νu 2 + r r r r r 1 1 ′ ′′ ⇒ u + u − 2 =0 r r 2 du 2d u ⇒ r +r −u=0 dr 2 dr Let's make a hange of variables
r = et ⇒ t = lnr du dt 1 du du = = dr dt dr rdt 1 d2 u du d 1 du d2 u = 2 = − dr 2 dr r dt r dt2 dt After substitutions:
r2
d2 u du +r −u=0 2 dr dr
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
114
d2 u du du − + −u=0 dt2 dt dt d2 u −u=0 ⇒ dt2 ⇒ u = C1 et + C2 e−t C2 ⇒ u = C1 r + r
⇒
The boundary onditions are
du u E = −P, kun r = R +ν = 1 − ν 2 dr r u(r = 0) < ∞ ⇒ C2 = 0 P (1 − ν) E (1 + ν)C ⇒ C = − ⇒ −P = 1 1 1 − ν2 E P (1 − ν) ⇒ u=− r E Nr0
The for e in the radial dire tion is thus
Nr0 =
E E (1 + ν)C1 = −P, Nϕ0 = (1 + ν)C1 = −P ⇒ Nr0 = Nϕ0 2 2 1−ν 1−ν
whi h do not depend on
r.
The potential energy expression is then
# ′ 2 Z " Z Z w 2ν ′′ 2 0 ′ 2 ′′ ′ ∆Π = πD (w ) + + w w rdr + π Nr (w ) rdr + π (ǫr Nr + ǫϕ Nϕ )rdr r r The last terms des ribes the energy due to the hanges of the midsurfa e. At the very moment of bu kling (w
6= 0) this term will vanish,
sin e the midsurfa e do not stret h,
ǫr = ǫϕ = 0. The
nal form of the total potential energy expression is thus
) # ′ 2 ZR ( " 2ν w + w ′′ w ′ − P (w ′)2 rdr D (w ′′ )2 + ∆Π = π r r 0
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
115
Example 5.5
Compute the riti al load of an isotropi ir ular plate ompressed uniformly in
the radial dire tion. Use a simple trial fun tion for the de e tion. Solution:
The potential energy expression is
) # ′ 2 ZR ( " 2ν w + w ′′ w ′ − P (w ′)2 rdr ∆Π = π D (w ′′ )2 + r r 0
Let's rst investigate a simply supported plate with a tral fun tion
w
as
r w0 r2 w = w0 1 − 2 ⇒ w ′ = −2w0 2 ⇒ w ′′ = −2 2 R R Z R r r 4r 3 r ⇒ ∆Π = π D 4 4 + 4 4 + 8ν 4 − P 4 drw02 R R R R 2 d ∆Π D = 0 ⇒ P = 4(1 + ν) 2 2 dw0 R The error is 24 % in omparison to the analyti al solution (Timoshenko, Gere: Theory of Elasti Stability). For a lamped plate let's use the trial fun tion:
w = w′ = w ′′ = ⇒ ∆Π = ⇒P =
The analyti al result is 14.68
2 r2 w0 1 − 2 R 2 r r 4w0 1− 2 − R R R r2 4w0 − 2 1−3 2 R R 32D 2 π − P w02 2 3R 3 D 16 2 R
D/R2 .
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
116
6
Non onservative problems, algorithms for linear algebrai eigenvalue problems
Example 6.1
Show, that the following load system is non- onservative. The load P remains
parallel to the beam A-B in the deformed on guration.
P 6
B
L q2 A b k2
? 6
L
q1
?O bk1
Solution:
The for e is onservative if there exists a potential fun tion
Fi = −
V
su h, that
∂V ∂qi
The virtual work done by su h loads is then
δW = Fi δqi = −
∂V δqi = −δV ∂qi
(63)
In our ase
δW = Pi δxBi P¯ = −P sin q2¯i − P cos q2¯j x¯B = (sin q1 + sin q2 )L¯i + (cos q1 + cos q2 )L¯j ⇒ δ¯ xB = (δq1 cos q1 + δq2 cos q2 )L¯i − (δq1 sin q1 + δq2 sin q2 )L¯j ⇒ δW = P L(− cos q1 sin q2 − sin q1 cos q2 )δq1 + 0 · δq2
From the equation below and from equation (63) we get
⇒
(
∂V ∂q1 ∂V ∂q2
= −P L sin(q1 + q2 ) = 0
whi h is learly a ontradi tion. Hen e
P~
⇒ V = V (q1 )
is not a onservative for e.
Non- onservativity of a for e an also be proven by examining the work done in a losed path. If su h a work will vanish for all possible losed paths (i.e. the work is path independent) the for e is onservati e. In the example it is easy to onstru t a losed path for whi h the work will not vanish. For example
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
117
1. rotating the bar OA
⇒ ∆W1 = 2P L
2. rotating the bar AB
180◦
180◦
ounter lo kwise keeping the bat AB in a verti al position
lo kwise
3. rotating the whole olumn OAB original upright position
⇒ ∆W2 = 0
180◦
lo kwise, after whi h the olumn is ba k in its
⇒ ∆W3 = 0
The total work done by this losed deformation path is
W = ∆W1 + ∆W2 + ∆W3 = 2P L 6= 0 ⇒
the for e
P~
is non- onservative.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
118
Example 6.2
Determine the riti al load of the rigid bar supported by a linear elasti spring
and a linear dashpot.
Solution:
P- b
k b
η
Sin e the bar is assumed to be rigid
b
! ! ϕ !
P
! !! ! 6 δ b!! - ? 6
A
EI = ∞. δ = Lϕ δ˙ = Lϕ˙ ZL/2 1 x2 dm = ρAL3 Jp = 2 12 0
1 1 JA = Jp + ρAL3 = ρAL3 4 3
F = η δ˙
The moment equilibrium with respe t to the point A gives
2 L ϕ¨ − kϕ − ηL2 ϕ˙ + P Lϕ = 0 −Jp ϕ¨ − ρAL 2 1 −(Jp + ρAL3 )ϕ¨ − ηL2 ϕ˙ − (k − P L)ϕ = 0 4 k − PL ηL2 ϕ˙ + ϕ = 0 ϕ¨ + JA JA Let's denote
ϕ¨ + aϕ˙ + bϕ = 0
and using a trial fun tion
ϕ = ert ,
we get the hara teristi
equation
a r 2 + ar + b = 0 ⇒ r = − 2
1±
r
4b 1− 2 a
!
The behaviour of the solution depends on the properties of the roots of the hara teristi equation. Let's assume the following initial onditions
ϕ(0) = ϕ0 , ϕ(0) ˙ =0 We have to investigate the following ve ases
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
119
1.
For
r1 , r2 ∈ R ja r1 < 0, r2 > 0 ϕ0 ⇒ ϕ(t) = r2 er1 t − r1 er2 t r2 − r1 ⇒ ϕ(t) → ∞, kun t → ∞ r2 > 0 , 1−
ϕ ϕ0
we have to have
4b k >1⇒P > 2 a L t
The solution is this unstable and behaves as shown in the following gure.
2.
r1 , r2 ∈ R ja r1 < 0, r2 = 0 4b k 1− 2 =1⇒P = a L ⇒ ϕ(t) = ϕ0
Now the equation is
3.
4.
ϕ ϕ0
t
ϕ¨ + aϕ˙ = 0.
r1 6= r2 ∈ R ja r1 , r2 < 0 4b ϕ 0 k/L.
In this ase the equilibrium is unstable. If we perturb the stru ture with a
small disturban e the bar starts to move towards the point B. The bar starts to vibrate around the point B and due to the vis ose damper the movement nally dies out and the bar remains in the tilted on guration. So the stru ture will not return to the original equilibrium state
ϕ = 0.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
121
Example 6.3
Determine the riti al load pf the two bar olumn using the dynami method.
The for e P is a dead-weight load, i.e. stays verti al.
P B?
6
L q2
b A k2 q1 O ? b k1 ? 6
L
Solution:
The equation of virtual work is
δWi + δWe = δWj X ¯ ¯ · δq δWi = Li δqi = L X ¯ ¯ · δq δWe = Qi δqi = Q X ¯ δWi = Ji δqi = J¯ · δq
Now
δWi = −M1 δq1 − M2 δq2
= −k1 q1 δq1 − k2 q2 δq2
δWe = P δv, jossa v = L[2 − cos q1 cos(q1 + q2 )] = P L [sin q1 + sin(q1 + q2 )]δq1 + sin(q1 + q2 )δq2 Z X x¨¯ · δ¯ xdρ, jossa dρ = mdξ δWj = sauvat
For bar 1:
ZL
x¨¯ · δ¯ xmdξ
0
x¯ = x¯˙ = δ¯ x = x¨¯ = ⇒
ZL 0
x¨¯ · δ¯ xmdξ =
sin q1 ξ cos q1 cos q1 ξ q˙1 − sin q1 cos q1 ξδq1 − sin q1 sin q1 cos q1 ξ(q˙1 )2 ξ q¨1 − cos q1 − sin q1 ZL 1 m ξ 2 q¨1 δq1 dξ = mL3 q¨1 δq1 3
0
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
122
For bar 2:
sin(q1 + q2 ) sin q1 ξ L+ x¯ = cos(q 1 + q2 ) cos q1 cos(q1 + q2 ) cos q1 ξ(q˙1 + q˙2 ) Lq˙1 + x¯˙ = 1 + q2 ) − sin(q − sin q1 cos(q1 + q2 ) sin q1 cos q1 2 ξ(¨ q1 + q¨2 ) ξ(q˙1 ) + L¨ q1 − x¨¯ = − sin(q1 + q2 ) − cos q1 sin q1 sin(q1 + q2 ) ξ(q˙1 + q˙2 )2 − cos(q1 + q2 ) ZL 1 1 3 ¨ x¯ · δ¯ xmdξ = mL q¨1 δq1 + cos q2 (¨ ⇒ q1 + q¨2 )δq1 − sin q2 (q˙1 + q˙2 )2 δq1 2 2
0
1 1 q1 + q¨2 )(δq1 + δq2 ) q1 (δq1 + δq2 ) + sin q2 (q˙1 )2 (δq1 + δq2 ) + (¨ − cos(2q1 + q2 )¨ 2 3
From the virtual work equation we get
¯+Q ¯ = J¯ L −k1 q1 ¯ L = −k2 q2 sin q + sin(q + q ) 1 1 2 ¯ = Q sin(q1 + q2 ) 1 1 4 q1 + q¨2 ) − sin q2 (q˙1 + q˙2 )2 J1 = mL3 q¨1 + cos q2 (¨ 3 2 2 1 1 1 2 − cos(2q1 + q2 )¨ q1 + sin q2 (q˙1 ) + (¨ q1 + q¨2 ) 2 3 3 1 1 1 3 2 J2 = mL − cos(2q1 + q2 )¨ q1 + sin q2 (q˙1 ) + (¨ q1 + q¨2 ) 2 3 3 Linearizing this non-linear equation at the point
(¯ qe , ¯0, ¯0)
we get
¯ ¯ ∂Q ¯ = Q(¯ ¯ qe , ¯0) + ∂ Q (¯ qe , ¯0)¯ q∗ + (¯ qe , ¯0)q¯˙∗ Q ˙ ∂ q¯ ∂ q¯ ¯ ¯ ∂ L ∂ L ¯ = L(¯ ¯ qe , ¯0) + L (¯ qe , ¯0)¯ q∗ + (¯ qe , ¯0)q¯˙∗ ∂ q¯ ∂ q¯˙ ¯ ¯ ¯ ¯ 0) ¯ + ∂ J (¯ ¯ 0)¯ ¯ q ∗ + ∂ J (¯ ¯ ¯0)q¯˙∗ + ∂ J (¯ J¯ = J¯(¯ qe , 0, qe , 0, qe , 0, qe , ¯0, ¯0)q¨¯∗ ∂ q¯ ∂ q¯˙ ∂ q¨¯ ¯ and Q ¯ do not depend on angular velo ities we have ∂ L/∂ ¯ q¯˙ = ∂ Q/∂ ¯ q¯˙ = 0. At equilibrium L ¯ 0) ¯ and also ∂ J¯/∂ q¯˙(¯ ¯ 0) ¯ = ∂ J¯/∂ q¯¨(¯ ¯ 0) ¯ = 0. qe , 0, qe , 0, qe , 0, have (¯
Sin e we
Sin e
¯ qe , ¯0) + L(¯ ¯ qe , ¯0) = J(¯ ¯ qe , ¯0, ¯0) Q(¯ we an write the linearized equation in the form
K q¯∗ + M q¨¯ = 0 Rak-54.131 Stability of stru tures exer ises / 12.12.2005
123
where
¯ ¯ ∂L ∂Q K = − (¯ qe , ¯0) − (¯ qe , ¯0) " ∂ q¯ # "∂ q¯ # k1 0 2 1 = − P L, attheequilibriumpoint 0 k2 1 1 " # 5 1 ∂ J¯ 1 M = mL3 (¯ qe , ¯0, ¯0) = ∂ q¯¨ 3 1 1
0 q¯e = 0
Using the trial fun tion:
Let's assume that
k1 = k2 = k
q¯∗ = est x¯ K + s2 M x¯ = ¯0
and denote
P = λk/L
and
r 2 = s2 mL3 .
Hen e we get
4 4 5 r + k 2 − λ r 2 + k 2 (1 − 3λ + λ2 ) = 0 9 3 q 2
⇒r =k
The riti al situation is when
r = 0,
−2 + 35 λ ∓
λ2 − 34 λ +
20 9
8/9
so we get
√ 5 3 λ − 3λ + 1 = 0 ⇒ λ = − 2 2 2
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
124
Example 6.4
Investigate the eigenvalue solution of the following stru ture dis retized by FEM. P
P
?
?
EI
EI
EI
e
2L
e
L
1. Show that in this parti ular ase the initial stiness matrix matrix
S
Solution:
K
and the geometri al stiness
are positive de nite.
A matrix is SPD (=Symmetri Positive De nite) if it is symmetri and all its
eigenvalues are positive. In this ase
4 −3
2
, S = K= −3 3 −3 2 −3 16
8 15 − 15 2 − 15
2 − 15 − 15 6 5 1 −5
− 15 8 15
The hara teristi equation is now third order polynomial, let's solve the eigenvalues by Matlab
>> K = [4 -3 2;-3 3 -3;2 -3 16℄ K = 4
-3
2
-3
3
-3
2
-3
16
>> S = [8./15 -0.2 -2./15; -.2 1.2 -.2; -2./15 -.2 8./15℄ S = Rak-54.131 Stability of stru tures exer ises / 12.12.2005
125
0.5333
-0.2000
-0.1333
-0.2000
1.2000
-0.2000
-0.1333
-0.2000
0.5333
>> k= eig(K) k = 0.3860 5.4268 17.1872 >> s = eig(S) s = 1.2899 0.6667 0.3101 2. Solve the lowest bu kling load and the orresponding eigenve tor by the inverse power iteration method
Solution: (a)
The inverse power iteration algorithm:
initial guess for eigenmode
(b)
iterating
φ1 ,
x1
whi h should have a omponent in the dire tion of the wanted
omputing the ve tor
k = 1, 2, ...,
until
y1 = Sx1
|(ρk+1 − ρk )/ρk+1 | < T OL Kxk+1 = yk yk = Sxk xTk+1 yk ρ(xk+1 ) = T xk+1 yk+1 yk+1 yk+1 = T (xk+1 yk+1 )1/2
If
y1T φ1 6= 0,
then
yk+1 → Sφ1 , ρ(xk+1 ) → λ1 , when k → ∞ Let's program this routine as a fun tion in Matlab
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
126
fun tion [arvo,vektori,tol,k℄=h19b(K,M,x1) k = 1; y1 = M*x1; p1 = 0; while (k > 0) xk =K\y1; yk = M*xk; pk = xk'*y1/(xk'*yk); tol(k) = abs((pk-p1)/pk); if (tol(k) < 1e-15) k = -k-1; end p1=pk; y1=yk/(xk'*yk)^.5; k = k+1; end eigenvalue = pk; eigenve tor = xk/(xk'*yk)^0.5; and running it gives
>> [eigenvalue eigenve tor tol℄=h19b(K,S,[1;1;1℄) eigenvalue = 0.5284
eigenve tor = 0.6820 0.9350 0.0823
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
127
tol = 3.8974e-09 3. Use shifting in the inverse power iteration, and use a shift whi h is (a)
50 %
(b)
90 %
( )
99 %
from the eigenvalue just solved.
Solution:
Shifting in uen es the rate of onvergen e of the inverse iteration. The idea is
the following:
Kφ = λSφ = aλSφ + (1 − a)λSφ ⇒ (K − µS)φ = ηSφ ˜ = ηSφ ⇒ Kφ
The eigenve tors are learly the same. The wanted eigenvalues an be obtained from
λ = η + µ.
The shifted inverse power iteration nds the lowest eigenvalue to the shift.
Using the Matlab fun tion above. Shifting is just substra tion from
K
the matrix
pλ1 S :
--------------- (a) --------------------->> sK = K - 0.5*oa*S; >> [a v t k℄=h19b(sK,S,[1;1;1℄) a = 0.2642
v = 0.6820 0.9350 0.0823
t =
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
128
8.9474e-08
k = -3 >> a+0.5*oa ans = 0.5284 --------------- (b) --------------------->> sK = K - 0.9*oa*S; >> [a v t k℄=h19b(sK,S,[1;1;1℄) a = 0.0528
v = 0.6820 0.9350 0.0823
t = 1.5519e-10
k =
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
129
-3 >> a+0.9*oa ans = 0.5284 --------------- ( ) --------------------->> sK = K - 0.99*oa*S; >> [a v t k℄=h19b(sK,S,[1;1;1℄) a = 0.0053
v = 0.6820 0.9350 0.0823
t = 3.6936e-07
k = -2 >> a+0.99*oa ans =
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
130
0.5284 4. Solve the same problem using the Rayleig quotien iteration
Solution: (a)
The Rayleigh quotient algorihm utilizes shifting at every iteration step:
starting from an initial quess of the vanted eigenve tor
(b)
iterating
k = 1, 2, ...,
φ1 ),
until
x1
(have to have a strong omponent in the dire tion
omputing ve tor
y1 = Sx1
|(ρk+1 − ρk )/ρk+1 | < T OL
(K − ρ(xk )S)xk+1 = yk
yk = Sxk xT yk ρ(xk+1 ) = T k+1 + ρ(xk ) xk+1 yk+1 yk+1 yk+1 = T (xk+1 yk+1)1/2
If
y1T φ1 6= 0,
then
yk+1 → Sφ1 , ρ(xk+1 ) → λ1 , when k → ∞ Let's program the RQI in a Matlab fun tion
fun tion [value,ve tor,tol,k℄=h19d(K,M,y1) k
= 1;
p1 = 0; sK = K; while (k>0) xk = sK\y1; yk = M*xk; pk = xk'*y1/(xk'*yk) + p1; tol(k) = abs(pk-p1); if (tol(k) < 1e-15) k = -k-1; end p1=pk; y1=yk/(xk'*yk)^.5; k = k+1; sK = (K-pk*M); end Rak-54.131 Stability of stru tures exer ises / 12.12.2005
131
value = pk; ve tor = xk/(xk'*yk)^0.5; and running it gives the solution
>> [a v t k℄=h19d(K,S,[1;1;1℄) a = 0.5284
v = 0.6820 0.9350 0.0823
t = 4.2295e-08
k = -3 5. Let's draw the onvergen e plots.
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
132
k¨a¨anteisiteraatio ⋄ shift + Rayleigh ◦
1+ ◦⋄
◦ + ⋄ ⋄ + ◦
+ ⋄
ln(virhe) 1e-10
+ ⋄
+ ⋄
◦
+ ⋄
1e-20 1
2
3
4 iteraatiokierros
Rak-54.131 Stability of stru tures exer ises / 12.12.2005
5
6
7
133
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