1.38 Temperature changes and displacement of supports

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To get things straight let us repeat the constitutive equations for a bar [0, l]

with mechanical and thermal loading, [115],

u

_ − ε = αT ΔT

EAε − N = 0 (1.520)

−N

_ = p .

u = uel + uT (1.521)

where uel = u − uT is the part that corresponds to the mechanical loading.

We assume that the bar is fixed on the left side, u(0) = 0, and that it carries

additionally a single force P at the other end at x = l. Green’s first identity

G(uel, v) = (p, v) + P · v(l) − a(uel, v) = 0 then implies

a(uel, v) =

_ l

0

p v dx + P · v(l) v ∈ V (1.522)

so that with uh

el = uh − uT

a(uh

el, ϕi) = a(uh − uT, ϕi) =

_ l

0

pϕi dx + P · ϕi(l) ϕi ∈ Vh (1.523)

or

a(uh, ϕi) = a(uT, ϕi) + fi ϕi ∈ Vh (1.524)

it follows

Ku = fT + f (1.525)

where

The longitudinal displacement of the bar is, see Fig. 1.134,

188 1 What are finite elements?

Fig. 1.134. Change of temperature and a displaced support

Fig. 1.135. Temperature changes should leave the bridge stress-free

k ij = a(ϕi, ϕj) fTi = a(uT, ϕi) fi =

_ l

0

pϕi dx + P · ϕi(l) .

(1.526)

In the following we employ a two-node bar element and we assume for simplicity

that the mechanical load is zero, p = 0, P = 0.

When the thermal loading is constant as in (1.520) then uT = αΔT x and

consequently the equivalent nodal forces are

fT1 = a(uT, ϕ1) = EA

_ l

0

u

_

T ϕ

_

1 dx = EA

_ l

0

αT ΔT ·

−1

l

dx

= −EAαT ΔT = −fT2 = −a(uT, ϕ2) . (1.527)

Hence the system

EA

l

_

1 −1

−1 1

__

u1

u2

_

=

_

fT1

fT2

_

(1.528)

has the solution u1 = 0, u2 = αT ΔT l and so the elastic part uel is—as we

expect—zero

uh

el = uh − uT = αT ΔT l · ϕ2(x) − αT ΔT x = αT ΔT(x

l

· l − x) = 0

(1.529)

1.38 Temperature changes and displacement of supports 189

Fig. 1.136. Change of temperature in one element only

which means that the bar is stress free σ = N/A = E ε = E (uh

el)_ = 0.

Next let the thermal loading increase with the distance x from the fixed

support, αT ΔT x, then

uT =

1

2 αT ΔT x2 (1.530)

and so

fT1 = −1

2EAαT ΔT l = −fT2 . (1.531)

Now the system (1.528) has the solution u1 = 0, u2 = 1/2 αT ΔT l2. But

because uh is linear and uT is quadratic the elastic solution is not zero

uh

el = uh − uT =

1

2 αT ΔT(x l − x2) (1.532)

and so spurious stresses σ = E (uh

el)_ appear in the bar which should be stressfree.

This means that temperature fields must have the same polynomial order

as the strains of the shape functions. If necessary higher order fields must be

interpolated by lower order functions: if the shape functions are quadratic the

temperature fields must be linear, if they are linear the temperature fields

must be constant, etc., see Fig. 1.135.

190 1 What are finite elements?

Example

In Fig. 1.136 the temperature increases in the second element only, so that

uT = αT ΔT x in the second element and uT = 0 in the first and the third

element. In these elements the elastic solution uel = uh − uT coincides with

uh and in the second element

uel = uh − uT = αT ΔT (

2

3 x − x) = −αT ΔT

x

3

(1.533)

so that N = −αT ΔT (EA/3) in that element.

Supports

If the support of a beam settles, w(l) = δ, the procedure is virtually the same.

The solution is

w(x) = wδ(x) + wel(x) , (1.534)

where wδ is a deflection curve with the property wδ(l) = δ and wel(x) corresponds

to the mechanical load p.

As before we have

a(wh

el, ϕi) = a(wh − wδ, ϕi) =

_ l

0

pϕi dx (1.535)

and so

Ku = fδ + f (1.536)

where

k ij = a(ϕi, ϕj) fδi = a(wδ, ϕi) fi =

_ l

0

pϕi dx . (1.537)

In both cases the equivalent nodal forces

fTi = a(uT, ϕi) = δWe(pT, ϕi) (1.538)

fδi = a(wδ, ϕi) = δWe(pδ, ϕi) , (1.539)

can be identified, via Green’s first identity

G(uT, ϕi) =

_ l

0

−EAu

__

T ϕi dx + [NT ϕi]l

0

_ _ _

δWe(pT ,ϕi)

−a(uT, ϕi) = 0, (1.540)

with the virtual work done by external forces pT and pδ, respectively. Note

that NT (x) = EAu_

T (x). In 1-D problems the loads pT are just the fixed end

forces ×(−1) due to the change in temperature αΔT,

1.38 Temperature changes and displacement of supports 191

i el, ϕi) = 0,

plied

Fig. 1.138. The FE solution wFE = vh

f + wδ

a(uT, ϕi) =

_ l

0

−EA(αΔT x)__

ϕi dx + [NT ϕi]l

0 = NT (l) ϕi(l) − NT (0) ϕi(0)

(1.541)

that is the forces (←→) the bar would exert on the confining walls if it were

fixed on both sides

← −f1 = NT (0) = EAαΔT → f2 = NT (l) = EAαΔT (1.542)

and the forces pδ result from the movement of the displaced node wi = δ,

because wδ is constructed by picking an appropriate nodal unit displacement

ϕi of the structure.

The solution technique can be summarized as follows:

• First all nodes are kept fixed, and the fixed end forces fTi and fδi due to

the temperature change or the movement of a node are calculated.

Fig. 1.137. Even in

load cases δ where

δW (w

projections are ap192

1 What are finite elements?

• The system Ku = f + fδ is solved, and with the nodal displacements ui

the elastic displacements and the elastic stresses are calculated.

• The stresses caused by the temperature change and the displaced support

respectively are added to the elastic stresses.

Projection

Also load cases δ are solved by a projection method, even when (seemingly)

p = 0. To see this, note that the strain energy product of the part wh

el of the

FE solution (1.534) with any ϕi ∈ Vh must be zero:

a(wh

el, ϕi) = a(wh, ϕi) − a(wδ, ϕi) = 0. (1.543)

To achieve this we proceed as follows: for the beam to assume the shape

wδ = δ · ϕe

3 (we let δ = 1), the nodal forces in Fig. 1.137 b must be applied.

These forces, so to speak, are the fixed end actions of the load case δ. Next

the opposite of these forces are applied to the original beam (see Fig. 1.137

c) i.e., when w(l) = 0. Let this load case be called −p0δ

and the associated

deflection curve vf . When this load case −p0δ

is solved with finite elements,

the deflection curve vf is projected onto the set Vh (see Fig. 1.138), and to

this function vh

f the deflection curve wδ is added. The result is the FE solution

wFE = wδ + vh

f .

Note that here—as in all standard 1-D problems—vh

f = vf because vf is

a piecewise third-order polynomial which lies in Vh.

Influence functions

If a support of a beam settles by 5 mm the Green’s function for the deflection

w(x) at any point x is

w(x) ·1 = R0(x) · 5mm (1.544)

where R0(x) is the support reaction due to the point load P = 1, the

Dirac delta δ0, acting at x, s. Sect. 7.3 p. 516. For any other quantity,

w_(x),M(x), V (x), R0 must be replaced by the appropriate support reaction

Ri corresponding to δi.

If the temperature in a frame element changes (αT = temperature strain)

or if an element is prestressed (N+) the influence function for the axial displacement

is

u(x) =

_ l

0

[N0 αT + ε0 N+ ] dx , (1.545)

where N0 and ε0 are the normal force and the strain respectively due to the

Dirac delta δ0. This result is based on a mixed formulation, see Sect. 4.19 p.

399, which provides the theoretical background for such problems. Though in

practice it is much simpler to think in terms of equivalent nodal forces and to

apply the negative end fixing forces to the structure and to follow their effects

with the standard influence functions for load cases p.

1.39 Stability problems 193

Fig. 1.139. Stability problem

and stress problem

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