Structures 4 lecture notes

University of Bath Department of Architecture & Civil Engineering

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Structures 4 lecture notes Buckling Buckling calculations are very difficult except for a few special cases, and so numerical methods on a computer are almost invariably used in practice for buckling modes involving plates, shells and assemblies of beams and columns. Single columns and beams aren’t too bad. The compression flange of a beam acts more or less like a column except that the torsional and warping resistance are involved. However the simple special cases are still worth examining because they tell us what to look for in a numerical analysis. The elastica

The sagging curvature is

dψ . This is the definition of curvature. The ds

sagging moment is equal to EI times the curvature. Sagging moment is also equal to −Py and thus


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dψ = −Py = −P ∫ sinψ ds ds s=0 s

M = EI

d 2ψ = −P sinψ ds 2 dψ d 2ψ dψ EI = −P sinψ 2 ds ds ds EI

1 ⎛ dψ ⎞ EI ⎜ ⎟ = P ( cosψ − cosψ 0 ) 2 ⎝ ds ⎠ 2

where ψ 0 is the value of ψ at s = 0 . Hence we have

1 ⎛ dψ ⎞ EI ⎜ ⎟ = P ( cosψ − cosψ 0 ) 2 ⎝ ds ⎠ 2

ψ

EI dψ s= ∫ 2P ψ =ψ 0 cosψ − cosψ 0

.

If we write ψ = 2θ , cosψ = cos2θ = 1 − 2sin 2 θ and ψ

s=

EI 2dθ = ∫ 2P ψ =ψ 0 2 sin 2 θ 0 − sin 2 θ

ψ

EI dθ 2 ∫ P sin θ 0 ψ =ψ 0 1 − k 2 sin 2 θ

1 k= sin θ 0

.

This is an incomplete elliptic integral of the first kind, - see http://en.wikipedia.org/wiki/Elliptic_integral This means that it cannot be evaluated using elementary functions – trigonometric functions, hyperbolic functions etc. Alternative derivation:


dy dx d 2 y dx d 2 y = 2 = cosψ dx ds dx 2 d2y d2y 2 dx 2 = dx3 = sec ψ 1 + tan 2 ψ

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tanψ = sec 2 ψ

dψ ds dψ ds

(

)

d2y dx 2

=

3 2

3

⎛ ⎛ dy ⎞ 2 ⎞ 2 ⎜ 1 + ⎜⎝ dx ⎟⎠ ⎟ ⎝ ⎠

Thus

EI

d2y dx 2

⎛ ⎛ dy ⎞ ⎞ ⎜ 1 + ⎜⎝ dx ⎟⎠ ⎟ ⎝ ⎠ 2

3 2

+ Py = 0

dy d 2 y EI dx dx 2 + Py dy = 0 3 dx ⎛ ⎛ dy ⎞ 2 ⎞ 2 ⎜ 1 + ⎜⎝ dx ⎟⎠ ⎟ ⎝ ⎠ EI 2

⎛ dy ⎞ 1+ ⎜ ⎟ ⎝ dx ⎠ 0

−

EI

1 + Py 2 = 0 2 ⎛ dy ⎞ 1+ ⎜ ⎟ ⎝ dx ⎠ 2

We can carry on, but we will get stuck again with an elliptic integral. However,

if

we

assume

1 ⎛ dy ⎞ ≈ ⎜ ⎟ 2 ≈1− 2 2 ⎝ dx ⎠ 1 ⎛ dy ⎞ ⎛ dy ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ 2 ⎝ dx ⎠ ⎝ dx ⎠ 1

1

This is satisfied by

2

that

dy dx

is

small,

⎛ ⎛ dy ⎞ 2 ⎛ dy ⎞ 2 ⎞ so that EI ⎜ ⎜ ⎟ − ⎜ ⎟ ⎟ + Py 2 = 0 . ⎝ ⎝ dx ⎠ ⎝ dx ⎠ 0 ⎠


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⎛ P ⎞ y = Bsin ⎜ x ⎝ EI ⎟⎠ ⎛ P ⎞ dy P =B cos ⎜ x . dx EI ⎝ EI ⎟⎠ P ⎛ dy ⎞ ⎜⎝ ⎟⎠ = B dx 0 EI This gives the Euler column formula,

π 2 EI π 2 EA = 2 L2 ⎛ L⎞ ⎜⎝ ⎟⎠ r A = cross-sectional area

P=

I = Ar 2

.

r = radius of gyration L = slenderness ratio r

For columns other than pin-ended columns, L is the effective length. For cantilever columns the effective length is more than twice the actual length, depending on how stiff the moment connection at the base is. Analysis of the elastica shows that if a column remains elastic the load continues to increase as the column buckles.


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The above figure shows a buckled pin ended column of length L and bending stiffness EI . The column is initially perfectly straight. The relationship between the buckling load and the shortening due to

π 2 EI P δ bending is where PEuler = 2 is the Euler buckling load. = 1+ L PEuler 2L This formula is obtained using complete elliptic integrals as described in §2.7 of Timoshenko and Gere, Theory of Elastic Stability and applies for small values of

δ . L

However, it can be shown that for the truss column below that the load

F decreases with deflection (see question 3 in the 2013-14 Structures 4 exam). This is because as the truss deflects the buckled member attracts more than its fair share of the load.


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Perry Robinson formula See the A history of the safety factors by Alasdair N. Beal, The Structural Engineer 89 (20) 18 October 2011, http://anbeal.co.uk/TSE2011HistoryofSafetyFactors.pdf for a fascinating discussion of safety factors including the Perry Robertson formula.

⎛ πx⎞ Assume column has an initial bend, y = ζ L sin ⎜ ⎟ (note that ζ is ⎝ L⎠ dimensionless) and that

dy is SMALL. Then the sagging moment, dx

⎛ d2y d2 ⎛ ⎛ π x⎞⎞⎞ M = EI × change of curvature = EI ⎜ 2 − 2 ⎜ ζ L sin ⎜ ⎟ ⎟ ⎟ = −Py ⎝ L ⎠⎠⎠ dx ⎝ ⎝ dx EI

d2y π2 ⎛ πx⎞ + Py = −EI ζ L sin ⎜ ⎟ ⎝ L⎠ dx 2 L2

⎛ πx⎞ Try solution y = Bsin ⎜ ⎟ then ⎝ L⎠

π2 ⎛ πx⎞ π2 ⎛ πx⎞ ⎛ πx⎞ −EIB 2 sin ⎜ ⎟ + PBsin ⎜ ⎟ = −EIζ L 2 sin ⎜ ⎟ ⎝ L⎠ ⎝ L⎠ ⎝ L⎠ L L π2 L2 = ζ L B= P π2 EI 2 − P 1 − 2 π EI L L2 EIζ L

The maximum stress is equal to

.


σ max

Z=

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PLζ P LAζ P M P PB P P Z = + = + = + = + A Z . P A Z A Z A 1− A 1− P 2 π EI PEuler 2 L

I is the section modulus. c

I Ar 2 P If we set I = Ar , Z = = σ = , σ max = σ y and c c A 2

σ Euler

PEuler π 2 EI π 2E = = = , then we have A AL2 ⎛ L ⎞ 2 ⎜⎝ ⎟⎠ r

σy =σ +

in which θ =

θσ . σ 1− σ Euler

LAζ Lcζ = 2 . Z r

Therefore

(σ Euler − σ )(σ y − σ ) + θσ Eulerσ = 0

(

)

σ 2 − σ y + (1+ θ )σ Euler σ + σ yσ Euler = 0

(

)

(σ

y

+ (1+ θ )σ Euler

)

(

)

(σ

y

+ (1+ θ )σ Euler

)

(

)

(σ

y

− (1+ θ )σ Euler

)

1 1 σ y + (1+ θ )σ Euler − 2 2 1 1 = σ y + (1+ θ )σ Euler − 2 2 1 1 = σ y + (1+ θ )σ Euler − 2 2

σ=

2

− 4σ yσ Euler

2

− 4 (1+ θ )σ yσ Euler + 4θσ yσ Euler .

2

+ 4θσ yσ Euler

When θ = 0 ,

(

)

1 1 σ y + σ Euler − σ y − σ Euler . 2 2 = σ y or σ Euler

σ=

.


Note that σ Euler = σ y when

π 2E L E =π . 2 = σ y so that r σy ⎛ L⎞ ⎜⎝ ⎟⎠ r

When σ Euler is very small,

(

)

(

)

1 1 σ y + (1+ θ )σ Euler − σ y2 − ( 2 − 2θ )σ yσ Euler 2 2 1 1 σ = σ y + (1+ θ )σ Euler − σ y 1− 2 (1− θ ) Euler 2 2 σy

σ!

!

⎤ 1 1 ⎡ σ σ y + (1+ θ )σ Euler − σ y ⎢1− (1− θ ) Euler ⎥ 2 2 ⎣ σy ⎦

(

)

! σ Euler and when σ Euler is very large,

σ!

(

)

1 1 σ y + (1+ θ )σ Euler − 2 2

2 − ( 2 − 2θ )σ yσ Euler (1+ θ )2 σ Euler

=

1 1 (1− θ ) σ y σ y + (1+ θ )σ Euler − (1+ θ )σ Euler 1− 2 2 2 (1+ θ )2 σ Euler

!

⎛ 1 1 (1− θ ) σ y ⎞ σ y + (1+ θ )σ Euler − (1+ θ )σ Euler ⎜ 1− 2 ⎟ 2 2 ⎝ (1+ θ ) σ Euler ⎠

( (

) )

1 ⎛ (1− θ ) ⎞ σ y = σ y ⎜ 1+ = 2 ⎝ (1+ θ ) ⎟⎠ 1+ θ

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Perry-Robertson graphs with

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E = 1000 and θ = 0, 0.01, 0.1 and 1 . σy

See http://en.wikipedia.org/wiki/Perry_Robertson_formula This is the basis for column design. Timoshenko or Cosserat column


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The figure on the left a shows the buckling of a battened or Vierendeel column. The deformation has been exaggerated so that the deflected shape can be seen. The figure on the right shows a detail of two bays in which the circles show the assumed points of contraflexure half way along the members. If the column is treated as a Timoshenko or Cosserat beam, the differential equations describing deformation of the column are

M = Py = −EI composite kθ = F = P

d ⎛ dy ⎞ ⎜⎝ − θ ⎟⎠ dx dx

dy dx

in which x is the vertical coordinate along the column, y is the lateral displacement of the column, M is the overall bending moment, P is the overall axial load and F is the overall shear force. I composite is the fully composite second moment of area and k is the shear stiffness of the column. Hence

EI composite

d ⎛ dy P dy ⎞ ⎜ − ⎟ + Py = 0 dx ⎝ dx k dx ⎠

⎛ P⎞ d y EI composite ⎜ 1− ⎟ 2 + Py = 0 ⎝ k ⎠ dx 2

.

If the column is pin-ended and its length is length L , the differential equation is satisfied by y = Bsin

if

πx L


π 2 EI composite ⎛ P ⎞ P= ⎜⎝ 1− ⎟⎠ L2 k P⎞ ⎛ = PEuler ⎜ 1− ⎟ ⎝ k⎠ P ⎞ ⎛ P ⎜ 1+ Euler ⎟ = PEuler ⎝ k ⎠ P=

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.

kPEuler k + PEuler

The shear stiffness k depends upon the bending stiffness and length of the individual members. In the above column all the horizontal members all have length 2a so that the length CF is a . The vertical members all have length 2b so that the lengths BC and CD are both b . The horizontal members all have second moment of area I horizontal

and the vertical members all have second moment of area I vertical and cross-sectional area Avertical . The members are all made from a material with Young’s modulus E .

ˆ is deformed from Shear deformation means that an angle such as DCF a right angle to

π + θ due to bending of the members. The 2

connections between the members are assumed to be rigid. If

we

ignore

I vertical ,

then

the

parallel

axis

theorem

gives

I composite = 2Averticala 2 .

To calculate the shear stiffness we first note that that the tip deflection of a cantilever of bending stiffness EI and span S loaded with a point

wS 3 load W at its tip is . The shear force in each of the vertical 3EI


members is 2

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F and the shear force in the horizontal members is 2

F b 2 = Fb . Thus the shear deformation is a a

⎛ F 3 ⎞ ⎛ Fb 3 ⎞ b a ⎜ 2 ⎟ ⎜ a ⎟ ⎜ 3EI vertical ⎟ ⎜ 3EI horizontal ⎟ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ Fb 2 Fab . θ= + = + b a 6EI vertical 3EI horizontal Hence k =

1 2

b ab + 6EI vertical 3EI horizontal

.

Buckling of plates and shells See, for example, Don O. Brush and Bo O. Almroth, Buckling of Bars,

Plates and Shells, McGraw-Hill, New York 1975. Plates behave fairly well when they buckle, the load usually does not drop off dramatically and may increase after buckling. On the other hand shells, including axially compressed cylinders, can be very imperfection sensitive so they collapse at a much smaller load than the eigenvalue buckling load - see Hunt, G. W., 2011. Reflections and symmetries in space and time. IMA

Journal of Applied Mathematics, 76 (1), pp. 2-26. http://imamat.oxfordjournals.org/content/76/1/2


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Simply supported flat plate

Image from Brush and Almroth, Buckling of Bars, Plates and Shells For a straight beam we have

d 2v + Pv = 0 dx 2 d 4v d 2v EI 4 + P 2 = 0 dx dx EI

where v is the displacement in the y direction. The corresponding equation for a flat plate is

⎛ ∂4 w ∂4 w ∂4 w ⎞ ∂2 w D⎜ 4 + 2 2 2 + 4 ⎟ +σ x 2 = 0 ∂x ∂y ∂y ⎠ ∂x ⎝ ∂x D=

Et 3 12 (1− υ 2 )

in which it is assumed that that there is only a membrane stress σ x in the x direction. Membrane stress has units force per unit width. w is the displacement in the z direction, t is the plate thickness and υ is Poisson’s ratio. D is the bending stiffness per unit width and it replaces the E

bd 3 for a rectangular beam. 12


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Let us suppose that the plate is simply supported along y = 0 and y = b and that it is long in the x direction. The differential equation is satisfied and the y = 0 and y = b boundary conditions are satisfied by w = Asin

2π x π y sin λ b

if 2 2 2 4 ⎛ ⎛ 2π ⎞ 4 ⎛ ⎛ 2π ⎞ 2 ⎛ π ⎞ 2 ⎞ ⎛ 2π ⎞ ⎛ 2π ⎞ ⎛ π ⎞ ⎛ π ⎞ ⎞ σ x ⎜ ⎟ = D⎜ ⎜ ⎟ + 2⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎟ = D⎜ ⎜ ⎟ + ⎜ ⎟ ⎟ ⎝ λ ⎠ ⎝ λ ⎠ ⎝ b⎠ ⎝ b⎠ ⎠ ⎝⎝ λ ⎠ ⎝⎝ λ ⎠ ⎝ b ⎠ ⎠

2

so that 2 ⎛ ⎛π⎞ ⎞ ⎜ ⎛ 2π ⎞ ⎜⎝ b ⎟⎠ ⎟ σ x = D⎜⎜ ⎟ + ⎟ ⎜ ⎝ λ ⎠ ⎛⎜ 2π ⎞⎟ ⎟ ⎜⎝ ⎝ λ ⎠ ⎟⎠

2

which is minimum when 2 2 ⎛ ⎛π⎞ ⎞⎛ ⎛π⎞ ⎞ ⎜⎝ ⎟⎠ ⎟ 2π ⎜ ⎛ 2π ⎞ ⎜⎝ b ⎟⎠ ⎟ ⎜ dσ x b = −2D ⎜ ⎜ ⎟ + =0 ⎟ ⎜ 1− 2⎟ . ⎝ ⎠ dλ ⎛ 2π ⎞ ⎟ ⎜ ⎛ 2π ⎞ ⎟ λ 2 ⎜ λ ⎜ ⎟ ⎜⎝ ⎝ λ ⎠ ⎟⎠ ⎜⎝ ⎜⎝ λ ⎟⎠ ⎟⎠

λ = 2b Thus the buckling membrane stress is

⎛π⎞ σ x = 4D ⎜ ⎟ . ⎝ b⎠ 2

Simple single degree of freedom models These will be introduced in lectures, with particular reference to postbuckled stability and imperfection sensitivity. In particular the


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difference between non-linear and linear (eigenvalue) buckling will be emphasised.

Linear or eigenvalue buckling

The above structure is loaded by a system of loads that are all multiplied by the same load factor λ . Imagine that it is analysed as a linear elastic structure with stiffness matrix

K and that the axial

forces in the members are found. These axial forces will all be proportional to λ and from these forces we can find the geometric stiffness matrix − λ G . The minus is put there because compressive forces produce a negative stiffness. Linear buckling occurs when

[K − λG]δ = 0 K −1 [ K − λ G ] δ = 0 ⎡⎣ K −1K − λ K −1G ⎤⎦ δ = 0 . ⎡⎣ I − λ K −1G ⎤⎦ δ = 0 1 ⎤ ⎡ −1 K G − I δ=0 ⎢⎣ λ ⎥⎦


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We are only interested in the lowest buckling load for which the load factor λ is equal to one over the highest eigenvalue of K −1G . The corresponding eigenvector gives the mode shape. Note the similarity to natural frequencies and mode shapes. Buckling corresponds to the natural frequency becoming zero. Note that linear eigenvalue buckling analysis gives no information about imperfection sensitivity. Therefore non-linear buckling analysis should always be done if there is any question of imperfection sensitivity.

The P − Δ effect refers to the moment caused by side-sway of a column. It is sometimes not clear what is meant by P − Δ analysis in a piece of software regarding linear or non-linear. Note that rotation is not a vector unless it is small and this assumption is often made even in so called non-linear analysis.


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Lagrange's Equations of Motion 1 n n 1 Kinetic energy = T = ∑ ∑ M ijδiδ j = δ T Mδ where n is the number of 2 i=1 j=1 2 degrees of freedom. The M ij are functions of the δ ’s only. n ∂ M ij ! ! ! ⎞ dT 1 n n ⎛ = ∑ ∑ ⎜ M ijδ!!iδ! j + M ijδ!iδ!!j + ∑ δ iδ jδ k ⎟ dt 2 i=1 j=1 ⎝ ⎠ k=1 ∂δ k

⎡ ⎛ ⎞⎤ ⎛ ∂ M jk ∂ M kj ⎞ + ⎢ n ⎜ M +M ⎥ ⎟ n ⎜ ∂δ i ⎠ ! ! ⎟ ⎥ 1 ⎝ ∂δ i ij ji !! ⎢ ! ⎜ ⎟ = ∑ δi ∑ δj + ∑ δ jδ k ⎢ j=1 ⎜ 2 2 k=1 2 ⎟⎥ i=1 ⎢ ⎜⎝ ⎟⎠ ⎥ ⎣ ⎦ n

(

)

⎡ ⎛ ⎞⎤ ⎛ ∂ M jk ∂ M kj ⎞ ⎛ ∂ M jk ∂ M kj ⎞ + + ⎢ ⎥ n n ⎜ M + M n ⎜ ∂δ i ⎟⎠ ! ! 1 n ⎜⎝ ∂δ i ∂δ i ⎟⎠ ! ! ⎟ ⎥ ⎝ ∂δ i ij ji !! ⎢ ! = ∑ δi ∑ ⎜ δj +∑ δ jδ k − ∑ δ jδ k ⎟ ⎢ j=1 ⎜ 2 2 2 k=1 2 ⎟⎥ i=1 k=1 ⎢ ⎜⎝ ⎟⎠ ⎥ ⎣ ⎦

(

)

n ⎡ ⎛ d ⎛ ∂T ⎞ ∂T ⎞ ⎤ = ∑ ⎢δ!i ⎜ ⎜ ! ⎟ − ⎟⎥ i=1 ⎢ ⎣ ⎝ dt ⎝ ∂δ i ⎠ ∂δ i ⎠ ⎥⎦

If U is the strain energy and G is the gravitational potential energy, by conservation of energy,

0=

n ⎡ ⎛ d ⎛ ∂ T ⎞ ∂ T ∂U ∂ G ⎞ ⎤ d (T + U + G ) = ∑ ⎢δi ⎜ ⎜  ⎟ − + + ⎟ ⎥ dt i=1 ⎢ ⎣ ⎝ dt ⎝ ∂δ i ⎠ ∂δ i ∂δ i ∂δ i ⎠ ⎦⎥

This equation applies for arbitrary δi and therefore d ⎛ ∂ T ⎞ ∂ T ∂U ∂ G − + + =0 . dt ⎜⎝ ∂δi ⎟⎠ ∂δ i ∂δ i ∂δ i

These are Lagrange’s equations of motion and apply for i = 1 to n .


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Verlet integration If the equations of motion can be rearranged so that:

( ) δ = q (δ , δ ,..., δ , δ , δ ,..., δ , p , p ,..., p ) δ1 = q1 δ1, δ 2 ,..., δ n , δ1, δ2 ,..., δn , p1, p2 ,..., pn 2

2

1

2

n

1

2

n

1

2

n

. . . .

(

δn = qn δ1, δ 2 ,..., δ n , δ1, δ2 ,..., δn , p1, p2 ,..., pn

)

in which the loads p1, p2 ,..., pn are known values of time and if we also know the initial values of δ1, δ 2 ,..., δ n , δ1, δ2 ,..., δn , then we can step through time updating values as follows:

δ1 = δ1 + δ1Δt δ = δ + δ Δt 2

2

δ1 = δ1 + δ1Δt δ = δ + δ Δt

2

. .

2

and

2

. .

.

.

.

.

δn = δn + δn Δt

2

.

δ n = δ n + δn Δt

There are a number of essentially similar ‘explicit’ methods like this with names such as Verlet Integration, Störmer's method, Gauss–Seidel and dynamic relaxation.


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Definitions of some terms used for nonaeroelastic vibrations Note that single degree of freedom systems are very important because multidegree of freedom systems can be reduced to uncoupled single degree of freedom systems using modal analysis and the orthogonality conditions discussed in lectures – this is ignoring the coupling due to damping. Summary of results In order to solve the equations ••

•

M δ + Dδ + Kδ = p for a system with n degrees of freedom, we write

n

δ = ∑ fr (t ) Δ r where r=1

Δ r are the eigenvectors of K −1M . If we ignore coupling due to damping, ••

•

mr fr + λr fr + kr fr = pr (t ) mr = Δ rT MΔ r kr = Δ rT KΔ r

λr = 2c kr mr pr (t ) = Δ rT p c = non-dimensional damping ratio which is the equation for a single degree of freedom system.


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Single degree of freedom system

Figure 1 Figure 1 shows a typical ‘random’ load, p(t ) .

1 The mean value of p(t ) is µ p = T

T 2

∫ p (t ) dt as T → ∞ .

−

T 2

The auto-correlation function is Rpp (τ ) =

1 T

T 2

∫ p (t ) p (t + τ ) dt as T → ∞ .

−

T 2

The auto-correlation function of p(t ) minus its mean is

1 T

T 2

∫ ⎡⎣ p (t ) − µ

p

T − 2

⎤⎦ ⎡⎣ p (t + τ ) − µ p ⎤⎦ dt as T → ∞

= Rpp (τ ) − µ p2 . The

φ pp (ω ) =

mean-square 1 2π

spectral

1 ⎡⎣ Rpp (τ ) − µ p2 ⎤⎦ e −iωτ dτ = −∞ 2π

∫

∞

∫

∞

−∞

density

⎡⎣ Rpp (τ ) − µ p2 ⎤⎦ cos (ωτ ) dτ

is .

Here

ω = 2π × frequency . From

the

theory

of

∞

∞

−∞

−∞

Fourier

Rpp (τ ) = µ p2 + ∫ φ pp (ω ) eiωτ dω = µ p2 + ∫ φ pp (ω ) cos (ωτ ) dω .

transforms,


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Note that µ p , R pp (τ ) and φ pp (ω ) are all real, and R pp (−τ ) = R pp (τ ) and

φ pp (−ω ) = φ pp (ω ) .

σ p is

σ p2 =

1 T

the

standard

T 2

∫ ⎡⎣ p (t ) − µ

−

T 2

p

deviation

of

p(t ) which

is

defined

as

⎤⎦ ⎡⎣ p (t ) − µ p ⎤⎦ dt as T → ∞ ∞

= Rpp ( 0 ) − µ p2 = ∫ φ pp (ω ) dω −∞

The mean-square spectral density tells us about the amount of ‘energy’ at different frequencies, but gives no information about the relative phase.

ψp

is

the

root

mean

square

(rms)

value

of

p(t )

and

ψ 2p = µ2p + σ 2p = R pp (0) . p(t)

x(t) m

s

λ

Figure 2 Figure 2 shows a mass - spring- damper. When the load p(t ) is applied,

d2x dx m 2 + λ + sx = p (t ) dt dt so that

p (t ) 1 d 2 x 2c dx + +x= 2 2 Ω dt Ω dt s


where Ω = c=

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s = 2π × natural frequency and the viscous damping factor, m

λ . 2 sm

The mean value of x (t ) is µ x =

µp . s

The mean-square spectral density of x (t ) is φ xx (ω ) =

Figure 3 shows plots of

1 2

⎛ ω ⎞ ⎛ 2cω ⎞ ⎜⎝ 1 − Ω 2 ⎟⎠ + ⎜⎝ Ω ⎟⎠ 2

2

.

⎛ φ pp (ω ) ⎞ ⎜⎝ s 2 ⎟⎠ 2

⎛ ω 2 ⎞ ⎛ 2cω ⎞ ⎜⎝ 1 − Ω 2 ⎟⎠ + ⎜⎝ Ω ⎟⎠

2

.


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Figure 3 The standard deviation of x (t ) is σ x =

∫

∞

−∞

φ xx (ω ) dω = Rxx ( 0 ) − µ x2 and if

c is small,

1 πΩ πΩφ Ω ( ) 1 1 pp 2π σx = = s 2c s ∞

∫

∞

−∞

⎡⎣ Rpp (τ ) − µ p2 ⎤⎦ cos (Ωτ ) dτ

2 1 Ω ∫−∞ ⎡⎣ Rpp (τ ) − µ p ⎤⎦ cos (Ωτ ) dτ = s 4c

where 2πΩ is the natural frequency.

2c


The ‘dynamic magnification factor’ is

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πΩφ pp (Ω ) . Note that this 2cσ p2

dynamic magnification factor applies only to the dynamic component of the load. Figure 4 shows the response to the load.

Figure 4 Note the dynamic magnification which can be seen by comparing the mean and standard deviation of the load and the response.

Multi-degree of freedom systems Let us suppose the ‘load’ exciting a particular mode of vibration is N

p (t ) = ∑ Ai qi (t ) where the qi (t ) are the pressures and the Ai are the i=1

associated areas times displacement in the mode (which may be positive or negative). N

The mean of the load exciting the mode is µ p = ∑ Ai µqi and the autoi=1

correlation function is


N

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N

Rpp (τ ) = ∑ ∑ Ai A j Rqiq j (τ ) i=1 j=1

N

N

= µ p2 + ∑ ∑ Ai A j ⎡⎣ Rqiq j (τ ) − µqi µq j ⎤⎦ i=1 j=1

where the cross-correlation,

1 Rqiq j (τ ) = Rq j qi ( −τ ) = T Again φ pp (ω ) =

1 2π

∫

∞

−∞

T 2

∫ q (t ) q (t + τ ) dt as T → ∞ . i

−

j

T 2

⎡⎣ Rpp (τ ) − µ p2 ⎤⎦ e −iωτ dτ .

Note that in general Rqiq j ( −τ ) ≠ Rqiq j (τ ) so that cross mean-square spectral

density,

φqiq j (ω ) = φqiq j ( −ω ) = φq j qi ( −ω ) =

1 2π

∫

∞

−∞

⎡ Rq q (τ ) − µq µq ⎤ e −iωτ dτ i j ⎦ ⎣ ij

will

be

a

complex function. However

in

doing

the

N

N

summation φ pp (ω ) = ∑ ∑ Ai A jφqiq j (ω ) ,

the

i=1 j=1

imaginary parts will cancel out. In terms of correlation functions, the response is given by

⎛ N N ⎞ ⎡ ⎤ A A R τ − µ µ ( ) ∑ ∑ i j ⎣ qi q j qi q j ⎦ ⎟ cos ( Ωτ ) d τ ∫ ⎜ ⎠ 1 −∞ ⎝ i=1 j=1 σx = . m 4c ∞

A note on Fourier series For our purposes a stochastic random load can be approximated by the Fourier series, ∞

p (t ) = µ p + ∑ n=1

(

(

2φ pp (ω n ) Δω 2 cos ω n t + β p (ω n )

))


where

ωn =

Page 26 of 28

2πn 2π and Δω = if the period, T , is sufficiently large. Stochastic means T T

‘governed by the laws of probability’. The autocorrelation function, ∞ ⎛⎡ µ + ∑ p ⎜⎢ 1 ⎣ n=1 Rpp (τ ) = ∫ ⎜ ∞ T T ⎜⎡ − µ + 2⎜⎢ p ∑ ⎝⎣ n=1 T 2

= µ p2 +

T 2

(

⎤ 2φ pp (ω n ) Δω 2 cos ω n t + β p (ω n ) ⎥ ⎦

))

(

2φ pp (ω n ) Δω 2 cos ω n (t + τ ) + β p (ω n )

( (

))

⎞ ⎟ ⎟ dt as T → ∞ ⎤⎟ ⎥⎟ ⎦⎠

∫ ∑ (2φ (ω ) Δω 2 cos (ω (t + τ ) + β (ω )) cos (ω t + β (ω ))) dt

1 T

∞

pp

−

n

n

p

n

n

p

n

T n=1 2

T 2

( (

) ) (

⎛ ⎛ cos (ω nτ ) cos2 ω n t + β p (ω n ) ⎜ 2φ pp (ω n ) Δω 2 ⎜ ∫T ∑ ⎜⎝ −sin (ω nτ ) sin ω n t + β p (ω n ) cos ω n t + β p (ω n ) n=1 ⎜ ⎝ −

1 =µ + T 2 p

∞

2

∞

(

= µ p2 + ∑ 2φ pp (ω n ) Δω cos (ω nτ ) n=1

∞

(

⎞⎞ ⎟ ⎟ dt ⎟⎠ ⎟⎠

)

)

)

= µ p2 + 2 ∑ φ pp (ω n ) cos (ω nτ ) Δω n=1

You can also use the Fourier transform, but this seems to cause problems with spectral densities unless you limit the time to

−

T T ≤t≤ . 2 2

Duhamel's integral Duhamel's integral is a bit like Verlet integration, except that it only applies to linear systems. The unloaded single degree of freedom system:

m

d2x dx + λ + sx = 0 2 dt dt

Ω= c=

s = 2π × natural frequency m

λ 2 sm

is satisfied by

p (t ) 1 d 2 x 2c dx + +x= 2 2 Ω dt Ω dt s

or and

the

viscous

damping

where factor,


( ((

x = e −cΩ t Asin

) )

1 − c 2 Ω t + B cos

Page 27 of 28

(( 1 − c ) Ω t )) . 2

If the mass is stationary and receives an impulse I at t = 0 , then when t>0,

I e x= m

−cΩ t

sin

(( 1 − c ) Ω t ) . 2

( 1 − c )Ω 2

Duhamel's integral treats the load as lots of little impulses so that

x=

t

1

m

( 1− c ) 2

∫e Ω

−cΩ (t−τ )

sin

0

(( 1 − c )Ω (t − τ )) p(τ ) dτ . 2

Seismic excitation

The ground motion in the above is y (note that horizontal ground motion is more of a problem than vertical). The equation of motion is

m x + λ ( x − y ) + s ( x − y ) = 0 . We can rewrite this as

m x + λ x + sx = λ y + sy so that the ‘load’ is λ y + sy . However it is more usual to write ••

•

m ( x − y )+ λ ( x − y )+ s ( x − y ) = −m y


Page 28 of 28

in which ( x − y ) is the motion relative to the ground and it is this motion which causes the stresses in the structure. The ‘load’ is now simply −m y . This is easy to implement in matrix notation for multidegree of freedom systems.

Aeroelasticity This is discussed in lectures using the example of the Fokker E.V (later the D-VIII) monoplane (divergence) and the Tacoma Narrows bridge (single degree of freedom non-classical flutter) - see http://books.google.co.uk/books?id=DnQOzYDJsm8C&dq=stall+flutt er+tacoma&source=gbs_navlinks_s Chris Williams

Structures 4 lecture notes

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