3.8 Sensitivity analysis

Back

In sensitivity analysis we want to predict how changes in a structure affect

the internal actions in a structure. The essential tool for this analysis are, as

we want to show, the Green’s functions.

If the stiffness of a member changes then the response of a structure will

change too, as in a spring, k u = f; see Fig. 3.36. The response of the spring

to the unit load f = 1 is G = 1/k and an increase in the stiffness, k + Δk,

changes the response to Gc = 1/(k +Δk) so that for any load f the response

before and after the change is

u =

1

k

f uc =

1

k + Δk

f . (3.171)

If we do a Taylor expansion of the updated “Green’s function”

1

k + Δk

=

1

k

− 1

k2 Δk + . . . (3.172)

then we better understand how a structure reacts to such changes

uc _

_

1

k

− 1

k

Δk

k

_

f = u − 1

k

Δk × _ _ u_

force

. (3.173)

314 3 Frames

Fig. 3.37. A drop in the bending stiffness in the central member will lead to an

increase in the bending moment in the leftmost member a) original bending moment

distribution M b) influence function G2 for M(x) c) bending moment M2 of G2

3.8 Sensitivity analysis 315

The stiffening of the spring, k → k +Δk, makes that the original u produces

too large a reaction (k + Δk) u = f + Δku in the spring and to annihilate

this excess force the opposite movement Δu = Δku/k (approximately) must

be superimposed on u.

In stiffness matrices the Taylor expansion has the following form, [75],

(K +ΔK)−1 = K

−1 −K

−1ΔKK

−1 + . . . (3.174)

and the analogue between (3.173) and

uc = (K +ΔK)−1f _ u −K

−1ΔKu (3.175)

is evident.

Let us apply some heuristic arguments to this equation

uc − u = −K

−1ΔKu (3.176)

and let us consider a beam with a change EI → EI +ΔEI. If we simply give

(3.176) an integral form then we obtain

wc(x) − w(x) = −

_ l

0

G0(y, x)

_ _ _

K−1

ΔEI

d4

dy4

_ _ _

ΔK

_ l

0

G0(y, z) p(z) dz

_ _ _

u

dy (3.177)

which even makes sense if the change EI → EI + ΔEI is uniform because

then

wc(x) − w(x) = −

_ l

0

G0(y, x) ΔEI

EI

p(y) dy (3.178)

that is a negative change, ΔEI < 0, is like an increase in the load that the

beam must carry. And a local change obviously means a local increase in p.

Single values

To trace the change of single values J(u) (= u(x), σ(x), . . .) we can adopt the

approach in Sect. 1.27 p. 113

J(uc) − J(u) = −d(u,Gc) = −d(uc,G) _ −d(u,G) (3.179)

where d(., .) is the symmetric term which we add to the strain energy product

of the structure, a(., .) + d(., .), to incorporate the change in the stiffness.

In the following we will consider some typical examples.

316 3 Frames

Cracks in a beam, EI → EI + ΔEI

If a single member, [x1, x2], in a frame cracks, EI → EI + ΔEI, then the

change in the bending moment M(x) at a point x—which must not lie on the

cracked element, it can be any point of the frame—is

Mc(x) −M(x) = −

_ x2

x1

ΔEI w

__

c G

__

2 dy = −

_ x2

x1

ΔEI

EI

McM2

EI

dy

(3.180)

where integration is done along the cracked member—only. The deflection

wc is the deflection of the cracked beam and the influence function G2 (=

deflection of the member under the action of the Dirac delta δ2 at x) is from

the uncracked frame. If the cracks are not too large we may approximate wc by

the original deflection w of the frame element. We tested this with the frame

in Fig. 3.37 where the bending stiffness in the central member, EI = 16, 296

kNm2, dropped by nearly 1/3 to EI + ΔEI = 11, 358 kNm2. According to

(3.180) the change of the bending moment M in the leftmost member should

be approximately

ΔM _ −ΔEI

EI

x

_ l

0

MM2

_ E_I _

ante

dy = 0.4 kNm (3.181)

while the true change is ΔM = 0.5 kNm.

Rule #1 : If the stiffness changes in a part [x1, x2] of a frame then the

change in any quantity ∂i w (w,w_,M, V ) at any point x is

∂i wc(x) − ∂i w(x) _ −

_ x2

x1

ΔEI

EI

MMi

EI

dx (3.182)

where Mi is the bending moment of the influence function Gi for ∂i w and M

is the bending moment due to the load p.

Change in an elastic support, k → k + Δk

If the stiffness of a spring changes, k → k+Δk, then the strain energy product

of a beam transforms as follows:

a(w, ˆ w) + kw(l) ˆ w(l) ⇒ a(w, ˆ w) + (k + Δk)w(l) ˆ w(l) (3.183)

so that in this case d(w, ˆ w) = Δk w(l) ˆ w(l) and we have:

Rule #2 : If the stiffness of a spring changes then the change in any

quantity ∂i w (w,w_,M, V ) at any point x is

∂i wc(x) − ∂i w(x) _ −Δk · Gi(l, x) · w(l) (3.184)

3.8 Sensitivity analysis 317

Fig. 3.38. Failure of support B and study of the changes in M(x)

where Gi is the movement in the spring due the action of the Dirac delta δi

(influence function for ∂i w) and w(l) is the movement of the spring due to

the load p.

Of course if the increase is too large, say if Δk→∞, then only the exact

formula

J(wc) − J(w) = −d(w,Gc) = −Δk · Gc(l, x) · w(l) = −RG · w(l)

(3.185)

can predict the change correctly. Note that RG = ∞·0 (or very nearly), which

is the support reaction in the spring due to the Dirac delta at x, tends to a

value corresponding to a fixed support.

Loss of a support

How we argue if the support of a structure fails—here the support B of the

beam in Fig. 3.38—may be illustrated by studying the consequences for the

bending momentM at the mid-point x. The influence function for the bending

moment M(x) is the response of the beam to the action of a Dirac delta, δ2

at x; see Fig. 3.38 b.

The support reaction RG due to δ2 equals the bending moment M(x) at x

if the support B settles by one unit length. This follows from Betti’s theorem

318 3 Frames

Fig. 3.39. The thoughtful engineer, [41]

(we ignore the actions on the left side at the fixed ends because they contribute

no work)

W1,2 = Ml w

_

Δ(x−) −Mr w

_

Δ(x+)

_ _ _

=0

+RG ·1 = +RG · 1

= M(x) · 1 + RΔ ·0 = W2,1 . (3.186)

The two moments Ml and Mr are of the same size but rotate in opposite

directions and because the slope w_

Δ(x) is continuous at x the work done by

the pair Ml and Mr is zero, so that indeed M(x) = RG.

Now if the support B is removed in the load case p then to compensate for

this loss the support reaction Rp must be applied in the opposite direction.

This will cause the deflection w(l) = Rp l3/(3 EI) and hence the change in

the bending moment is

ΔM = −RG · w(l) = −RG · Rp _ _ _

ante

· l3

3 EI

← 1

kS

. (3.187)

Rule #3 : If a support fails then the change in any quantity ∂i w

(w,w_,M, V ) at any point x is

∂i wc(x) − ∂i w(x) = −RG · Rp · 1

kS

(3.188)

where RG is the support reaction due the action of the Dirac delta δi (influence

function for ∂i w), Rp is the support reaction due to the load p and kS is the

stiffness of the structure in the direction of the missing support.

Let us apply this result to the problem of the thoughtful engineer which

Galileo mentions in his discorsi, [41]: Il mecanico places a heavy marble column

on an additional support (see Fig. 3.39). But in so doing he unwillingly

lifts the left end of the column from its support and so instead of lowering the

3.8 Sensitivity analysis 319

bending moment by a factor of four as intended his maneuver effects only a

change in the sign of the bending moment M = Mmax.

Because if the weight of the column is g = 10 kN/m and its length l = 8

m then

M =

10 · 82

8

= 80

_ _ _

2 supp.

M = −10 · 42

8

= −20

_ _ _

3 supp.

M = −1

2

10 · 42 = −80

_ _ _

cantilever

(3.189)

so nothing is gained.

In the transition from the 2-span beam to the cantilever the change of the

bending moment is according to (3.188)

Mc(x) −M(x) = −RG · Rp · 1

kS

= −8, 480 · 15 · 1

2, 120

= −60kNm

(3.190)

which agrees with (3.189). A point load P = 1 kN at the end of the cantilever

beam effects the deflection 0.4708 mm and so kS = 1/0.4708 · 103 = 2, 120

kN/m. The force RG = 8, 480 is the support reaction in the 2-span beam from

the Dirac delta δ2 (≡ tan ϕ = 1).

If a rigid support yields

If a rigid support turns soft (∞ → kS) the logic is the same. It is only that

the end of the beam is now working against the bending stiffness of the beam

(3 EI/l3) and the soft support with its relic stiffness kS so that—the two

springs are working in parallel—the resulting stiffness is

k =

3 EI

l3 + kS (3.191)

and consequently

ΔM = −RG · w(l) = −RG · Rp _ _ _

ante

· 1

k

(3.192)

where RG and Rp have the same meaning as before. It is evident that this logic

applies to all types of supports and to internal nodes as well. If a previously

fixed support begins to rotate then (3.192) would have to be replaced by

ΔM = −MG · ϕ(l) = −MG ·Mp _ _ _

ante

· 1

kϕ

(3.193)

where MG is the fixed end reaction in the load case δi and Mp is the fixed end

reaction in the load case p and kϕ is the rotational stiffness of the support.

320 3 Frames

Fig. 3.40. Loss of a member—the energy needed to bend a spline into the shape

wc in the second span is indicative of the effect the loss will have on the structure

a) deflection w and wc b) bending moment M0 of the Green’s function G0

Rule #4 : If a rigid support softens then the change in any quantity ∂i w

(w,w_,M, V ) at any point x is

∂i wc(x) − ∂i w(x) = −RG · Rp · 1

k

(3.194)

where RG is the support reaction due the action of the Dirac delta δi (influence

function for ∂i w), Rp is the support reaction due to the load p and k is the

relic stiffness of the structure in the direction of the support.

Loss of a frame member

A complete loss of a frame member corresponds to EI = 0 or in terms of the

equation EI +ΔEI = 0 means ΔEI = −EI so that in the end

− d(wc,G) = −

_ l

0

ΔEI w

__

c G

__

dy =

_ l

0

EI w

__

c G

__

dy (3.195)

where wc is the shape of the member when it has lost all the stiffness; see Fig.

3.40 a. Its curvature −w__

c is the quotient of the bending moment Mc in the

member and the residual stiffness EI +ΔEI

3.8 Sensitivity analysis 321

− w

__

c= lim

ΔEI→−EI

Mc

EI +ΔEI

≡ 0

0 . (3.196)

The deflection wc in the frame member can be found as follows: remove the

member and let the frame find its new equilibrium position. Next bridge the

gap by a spline which attaches seamlessly to the two displaced nodes on both

sides of the gap and which has the stiffness EI. This is correct, not EI = 0,

because we need the bending moment Mc = −EI w__

c in the second equation

(3.195). Also the spline is not really attached to the nodes, rather prestressing

forces at the ends of the spline keep the element in the shape wc. If we would

reattach the member and release the prestressing forces in the member then

the structure immediately would snap back into its original shape.

The effect EI = 0 is cared for by releasing the two nodes that is by letting

the structure find its equilibrium position without the member. After that

we calculate how much the member must be prestressed to position its ends

opposite to the displaced nodes of the structure.

• The energy needed for this maneuver is an indication of how important

the element is for the structure that is how much J(w) will change if the

member fails.

The two-span beam in Fig. 3.40 loses its member in the second span. How

much will this affect the deflection at the center of the first span? The rotation

of the beam at the end of the first span is w_

c(4.0) = −0.00018 so that the

spline wc in the next span is the solution of the following problem

EI wIV

c = 0 wc(4.0) = wc(8.0) = Mc(8.0) = 0 w

_

c(4.0) = −0.00018 .

(3.197)

The bending moment distribution of this spline is linear, Mc(4.0) = −12.24

kNm and Mc(8.0) = 0, so that according to (3.195)

wc(2.0) − w(2.0) =

_ 8

4

EI w

__

c G

__

0 dy =

_ 8

4

McM0

EI

dy

=

1

3

(−0.21) · (−12.24) · 4.0 · 1

90, 480

= 0.04mm

(3.198)

which agrees with the exact result

wc(2.0) − w(2.0) = 0.18mm − 0.14mm = 0.04mm. (3.199)

Of course (3.195) is purely theoretical because the spline wc requires the

calculation of the displacements and rotations of the two neighboring nodes

after the member has been removed but then one can compare the two systems

directly.

322 3 Frames

Fig. 3.41. Release of the connection between two frames. The energy needed to

reconnect the spline wc with the two released nodes is an indication of how much

influence the stiff element exerts on the structure

The idea to substitute for wc _ w the original deflection would be in most

cases too crude an approximation—here 0.02 mm instead of the exact 0.04

mm—but it could suffice to signal the trend into which direction things will

be moving if the member fails.

Note that

|d(wc,G)| ≤ d(wc, wc) · d(G,G) =

_ l

0

EI (w

__

c )2 dx ·

_ l

0

EI (G

__)2 dx

(3.200)

so that the product of the strain energy of the spline wc and the Green’s

function G in the member is an upper bound for the change J(wc)−J(w). If

either of these two terms is small the change will not be very pronounced.

On the other hand imagine a short bolt which ties two structures together

and forces the structures to move in unison; see Fig. 3.41. Assume this bolt

fails—technically a very stiff, very short frame element—then the released

nodes will probably undergo large displacements and rotations and so the

spline wc which later must reconnect with the two released nodes will have

to assume a serpent like shape and to stretch a long way. Consequently the

3.8 Sensitivity analysis 323

Fig. 3.42. A prestressed member a) is placed under the end of the cantilever beam

and b) then released. This is the same logic as in the case of a loss of a member—only

in the other direction. If the prestressing forces are released the element assumes its

original shape c) and the gap δ10 must be closed d) by the redundant X1 = −δ10/δ11.

strain energy in this very stiff spline will be quite large, that is the change

(3.195) will no longer be negligible.

Rule #5 : If a member [x1, x2] fails, EI → 0, then the change in any

quantity ∂i w (w,w_,M, V ) at any point x is

∂i wc(x) − ∂i w(x) =

_ x2

x1

EI w

__

c G

__

i dy (3.201)

where wc is the shape the member assumes if it is drained of all its stiffness

and Gi is the influence function for ∂i w.

Adding a member to a structure

All this can be applied in the opposite direction as well: if the end of a cantilever

beam is placed on a vertical pin-jointed frame element (see Fig. 3.42)

324 3 Frames

Fig. 3.43. Footprint of a Green’s function or sensitivity map: which changes do

influence M(x) the most? EI = 90, 625 kNm2, members 0.5 ・ 0.3 ・ 6.0m

this can be done in two ways: (1) the length of the new frame element is

h − w(l) where h is the height of the cantilever beam before the load was

applied or (2) the length of the new element is h. In the first case the beam

would only benefit from the support when it carries additional loads. In the

second case the element first must be compressed so that it fits under the

beam—this would be the shape uc (in axial direction). Next the end of the

element is released so that it presses the cantilever beam upward. The change

in any quantity of the beam would then be

J(ew) = −

_ l

0

EAG

_

i u

_

c dy = −

_ l

0

Ni Nc

EA

dy (3.202)

where Ni is the normal force in the frame element due to the Green’s function

for ∂iw and Nc is the prestressing force. Not surprisingly all quantities

wc, w_

c,Mc, Vc would be just as large as if the additional support had been

3.8 Sensitivity analysis 325

present from the start. (In some sense the technique to prestress an element—

so that it fits into the statically determinate structure—and then to release it

is the force method in disguise).

So we better understand the role of the splines wc in Fig. 3.195 and Fig.

3.41. They act like batteries, they provide the energy to restore the original

shape of the structure. By removing an element from the frame the strain

energy of the structure increases which means that additional energy can be

gained from the sagging load. To restore the original shape we must give this

energy back to the structure, that is the load must be pushed up to its previous

level. The energy needed for this maneuver comes from the prestressing forces

in the splines.

The case of the missing member

If in the construction phase of a structure a member has been omitted unwillingly

the defect can be amended by prestressing the member so that it fits

into the deformed structure and by then releasing the prestressing force. The

shape of the structure is then the same as if the member would have been

present from the beginning.

We follow a similar procedure when we replace structural members, say,

the rusty old pier of a bridge. When the hydraulic jacks on top of the new

pier have lifted the bridge just enough to release the old pier the new pier

is prestressed by precisely the right amount to replace the old pier without

affecting the stress distribution in the bridge.

The important point is, of course, that the unstressed member has the

correct length. So eventually the sequence of the single construction stages

must be taken into account.

Summary

A change in any quantity J(u) → J(u)+ΔJ(u) means an increase or decrease

in the Dirac energy. In the case of a support this means

ΔJ(u) · 1 = force · displacement increment . (3.203)

The force is the support reaction RG due to the Dirac delta and the displacement

is the incremental (additional) movement of the support induced by the

change of the stiffness. So if the support of a node fails completely (or partially)

and the load p presses the node downward by w(xi) additional units

then the change is RG · w(xi).

In (3.187) the original movement is zero (rigid support). With the loss of

the support the system becomes a cantilever beam which deflects at its end

by Rp/k units; this is the deflection increment.

In (3.192) the original movement is zero too. But this time the support

does not yield completely and so Rp must work against the stiffness of the

326 3 Frames

node plus the relic stiffness of the support so that the deflection increment

will be somewhat lower.

The sensitivities in a structure are encapsulated in the “footprints” of the

Green’s functions; see Fig. 3.43. In this figure are plotted the bending moment

distribution (M2) and the support reactions of the influence function

for M(x), which is the bending moment at the center of the upper left horizontal

member. From this figure we can learn for example that if the central

support yields by 1mm in vertical direction then the change inM(x) is −1, 590

kN·0.001 m = - 1.59 kNm. Or a slip of 0.001 m in horizontal direction would

effect a change of 233 · 0.001 m = 0.23 kNm. And a rotation tan ϕ = 0.001 of

the support would change M(x) by −303 · 0.001 = -0.3 kNm.

But also the bending moments M2 of the influence function tell a story. If

at the upper left node, on the side A of the node, the concrete cracks—so that

the fixed end begins to resemble a rotational spring—then for any differential

degree ϕ of rotation between the horizontal member and the vertical member

the change in M(x) is 8, 237kNm · tan ϕ.

Safety of structures

So the failure of a support is critical for the safety of a structure if (1) the

support reaction Rp is large because then the incremental movement w(l) will

be large too (probably) and if (2) the support reaction RG in the load case

δi (= influence function for the internal actions M(x) or V (x) at x) is large,

see (3.187).

Hence, to assess the safety of a structure we could adopt the following

strategy: (1) Find the points x where the internal actions, M(x), V (x), etc.

attain their maximum values. (2) Calculate the influence functions for M(x)

and V (x). (3) Study how changes in the stiffness of the structure would influence

the distribution of the influence function and therewith the maximum

values of M(x) and V (x). The same can be done with the support reactions.

Of course a seasoned structural engineer needs no computer to see where

the weak points of a structure are but this is not the point here. Rather

this technique eventually could allow us to assess the safety of a structure

computationally.

4. Plane problems

We start with an elementary example to explain the FE technique in detail.

Навигация