ContMech SoSe2013 Exercise04 Solution

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Institute for Theoretical Physics Heidelberg University

Prof. Dr. Ulrich Schwarz Dr. Irina SurovtsovaContinuum mechanics Summer term 2013

Exercise sheet 4

Handout 13/05/2013—Return 27/05/2013—Discussion 03/06/2013

Exercise 4.1 [6 Punkte]: Muscle fibers

Consider a system plotted on the figure. An active motor force is F m = F 0 (1− v/v0). We identify

v = −u ( u is the displacement). The resting length is denoted by L0 and the external tension by F .

1. Derive the constitutive equation.

2. Consider the following length-controled experiment: the fiber is extended to L = 3L0

. At time

t = 0, this is reduced to L = 2L0. Calculate the time course for F . Under which condition on

F 0 and C 3 this solution corresponds to the behaviour of real muscle?

Solution

F = F 0

1 +

ε2v0

+ ηε2

ε2 = ε− F

C 3

where ε is the full strain, ε2 - strain in the dashpot.

Hence

F = F 0

1 +

ε− F /C 3v0

+ η

ε− F

C 3

The constitutive equation:

F

F 0

C 3v0+

η

C 3

+ F − F 0 = ε

η +

F 0C 3v0

Denoting

λ =F 0

C 3v0+

η

C 3, β = η +

F 0C 3v0

1



we get the following ODE:

λF + F − F 0 = εβ

with the solution:

F (t) = F 0 + (F (0+)− F 0) exp(−t/λ)

F (∞) = F 0 have to be < F (0−) and (F (0+) − F 0) < 0 to get increasing time course.

Hence, in order to get the solution for behavior of real muscle we need the following condition:

F (0+) < F 0 < F (0−)

Initial value of F (0+)

Using the differential equation:

(F (0+)− F (0−))λ = β (ε(0+)− ε(0−)) = −β

F (0

−

) =

3L0

−L0

L0 C 3 = 2C 3

F (0+) = −β

λ+ 2C 3 = C 3

The condition on F 0 and C 3C 3 < F 0 < 2C 3

Exercise 4.2 [6 Punkte]: Coarse-graining a viscoelastic chain

Consider a chain of N Kelvin-Voigt (Maxwell) units of the length a as shown in the figure. Using the

continuum limit ( aL

= 1N → 0) derive the stress-strain relations:

Kelvin-Voigt: F = Eε + ηdε

dt

Maxwell:dε

dt=

F

η+

1

E

dF

dt

2



Solution:

Kelvin-Voigt model

In the each Kelvin-Voigt unit i the strain ε experienced by the spring is the same as that experienced

by the dash-pot. Hence:

F i = Eε + ηdε

dt

where the strain εi in the i−th element is

εi =a

a

The forces are equal:

F = F 1 = ... = F N

The strain of the complete chain is

ε =L

L=

L

a · N =a1 + ... +aN

a ·N =

ε1 + ... + εN N

Hence we have for the strain rate:

ε =1

N ( ε1+...+ ˙εN ) =

1

N

F − Eε1

η+ ... +

F −EεN η

=

1

N

N

F

η− ... − E

η(ε1 + ... + εN )

Leading to the constitutive equation:

ε =F

η− E

ηε

Maxwell model

Here again the forces in series are equal:

F = F 1 = ... = F N

For each element we have

εi =F

η+

1

E ηF

and

εi =a

a

The strain in the chain is

ε =ε1 + ... + εN

N

Similar toKelvin-Voigt model:

ε =1

N

N

E F +

N

ηF

And finally:

dε

dt=

F

η+

1

E

dF

dt

3



Exercise 4.3 [6 Punkte]: Two contractile bars on an elastic foundation

Consider two contractile 1d elastic bars connected to an elastic

foundation with spring constant kf . σ0

denotes the internal con-

tractile stress (force dipoles). The bars are elastically coupled by

a linear spring of stiffness kc, leading to

σ(1)|x=L − σ(2)|x=L = kc

u(1)|x=L − u(2)|x=L

,

where u(i) is the displacement in ith bar. The layer’s ends (x = 0, 2L) are stress-free.

1. Find the solution for stresses σ(i)(i = 1, 2) in each layer.

2. Discuss the stress distribution in the bars for two cases:

a) weak coupling between bars: kc kf .

b) strong coupling between bars: kf kc.

Solution

Consider the displacement ui, i = 1, 2 in each bars. The governing equation are (see lecture 6.05.2013):

l2 pd2ui

dx2− ui = 0, i = 1, 2

The localization length l p = C kf

is the same for each bar.

We looking for the solution of this equations in the form (symmetrical with respect to x = L):

u1

(x) = A1

cosh xlp

+ B1

sinh xlp

u2(x) = A2 coshx−2Llp

+ B2 sinh

x−2Llp

The corresponding stresses are:

σi(x) = C dui

dx+ σ0

The four constants Ai, Bi have to be defined from the 4 boundary conditions:

σ1(x = 0) = σ2(x = 2L) = 0

σ(1)

|x=L = σ(2)

|x=L = kc u(1)

|x=L

−u(2)

|x=L ,

From the boundary condition at x = 0, 2L follows:

B1 = B2 = −σ0kf

For constants A1 and A2, we get after some calculations:

A2 = −A2 =(a1 + b1)q

det

where a1 = kf sinh Llp− kc cosh L

lp; b1 = kc cosh L

lp; q = 2kcB1 sinh L

lp− σ0(1− cosh L

lp)

4



For weak coupling kc kf , the absolute values of internal stresses are maximal at the center of each

bar and negligible at the beam-beam junction.

For a strongly coupled bars, kf

kc, internal stresses build up at the junction between the beams:

Here is chosen L = 10, l p = 1, σ0 = −50;

kf = 1, kc = 0.1 on the left graph and kf = 0.2, kc = 1 on the right one

Exercise 4.4 [4 Punkte]: Geometrical interpretation of the Green-Lagrange strain tensor

1. Consider a material line segment dξ 1. Derive the elongation of the segment after deformation

E ξ1 in terms of the Green-Lagrange strain tensor if the segment is deformed parallel to the

initial state. Discuss the linear approximation.

2.Consider now the shear deformation: the 900 angle between unit

vectors e1 and e2 deformed to angle ϕ.

Define the new unit vectors e∗1

and e∗

2and calculate sin ϕ = cos(e∗

1, e∗

2) in terms of the Green-

Lagrange strain tensor. Discuss the linear approximation.

Solution

The Green-Lagrange strain tensor:

E =1

2(F T F − 1)

1. A geometrical meaning of the normal components of Green-Lagrange strains E 11, E 22 and

E 33 is provided by considering the length and angle changes that result from deformation. We

consider a material line element dξ 1 that deforms to the element dx1. The orientation does not

change after deformation The stretch ratio is (see lecture):

λ =dξ 1dx1

The Green-Lagrange strain is defined by:

εGL =λ2 − 1

2=

1

2 e0 · (F T F − 1) · e0 = e0 · E · e0

5



For our case:

e0 = 10

0

This leads to:λ2 − 1

2= E 11, λ =

2E 11 + 1

Hence extension of the element (relative change of length) is:

E ξ1 :=dx1 − dξ 1

dξ 1=

2E 11 + 1 − 1

For linear case:

E ξ1≈

E 11

2. The geometrical meaning of the shear strain E 12 is found by considering the angles between

two directions e1 and e2 The unit vectors after deformation are defined:

e1∗ =

F · e1 e1 · F T F · e1

, e2∗ =

F · e2 e2 · F T F · e2

The cosinus between the deformed vectors is defined by scalar product:

cos φ = e1∗ · e2

∗ =F · e1

e1

·F T F

· e1

· F · e2

e2

·F T F

· e2

= e2 · F T F · e1

e1

·F T F

· e1 e2

·F T F

· e2

Using

e1 =

1

00

e2 =

0

10

we get e1 · F T F · e1 =

2E 11 + 1,

e2 · F T F · e2 =

2E 22 + 1

And finally (remember e1 and e2 are orthogonal for deformation ( e1 · e2 = 0)

cos φ = 2 e2 · (2E + 1) · e1

√ 2E 11 + 1√ 2E 22 + 1

= 2E 21

√ 2E 11 + 1√ 2E 22 + 1For small deformation:

E 21 ≈ 1

2cos φ

6

ContMech SoSe2013 Exercise04 Solution

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