Download Suggested solutions to 2015 MEK2500 Mock Exam

Suggested solutions to 2015 MEK2500 Mock
Exam
November 25, 2015
Oppgave 1
a) The displacement (in Lagrange representation) is:
u(x) = f (x) − x
(1)
Then Cauchy’s strain tensor is:
(2)
ε=
1
1
(∇f − ∇x) + (∇f − ∇x)T =
∇f + ∇f T − I
2
2
where I is the identity matrix.
b) Strain is dimensionless, thus has no unit.
c) Stress is defined as force per unit area with SI (derived) unit Pascal (Pa).
d) The tensile strength of an elastic material is the maximal tensile stress
(tension) the material can sustain before failure.
e) Natural materials have ν ∈ (0, 0.5). If a pure pressure is applied to the
outer boundaries of a body with ν > 0.5, the material will expand (in
contrast to contract as normally expected)
Oppgave 2
Assume that an elastic body B occupies a domain Ω ⊂ R3 with coordinates
x = (x1 , x2 , x3 ). Assume that the body is deformed with a deformation f
given by:
(3)
f (x) = ((κ1 + 1)x1 , (κ2 + 1)x2 , x3 )
1
for real, positive constants κ1 , κ2 . Let the Frobenius norm of an n × n matrix
A be denoted kAk with
2
kAk =
(4)
n X
n
X
a2ij
i=1 j=1
a) (5 points) Compute the inverse deformation g = f −1 that maps a coordinate y ∈ f (Ω) to x ∈ Ω.
We know that
(5)
(y1 , y2 , y3 ) = ((κ1 + 1)x1 , (κ2 + 1)x2 , x3 )
The inverse deformation g which maps y to x is then
(6)
x1 =
1
y1 ,
(κ1 + 1)
x2 =
1
y2 ,
(κ2 + 1)
x3 = y3 .
or, in other words
(7)
g(y) = (
1
1
y1 ,
y2 , y3 )
(κ1 + 1)
(κ2 + 1)
b) (5 points) Compute the displacement u = u(x) of the body B and
Cauchy’s infinitesimal strain tensor ε associated with u.
The displacement u is:
(8)
u = f − x = (κ1 x1 , κ2 x2 , 0)
The displacement gradient is:

(9)

κ1 0 0
∇u =  0 κ2 0 
0 0 0
Since ∇u is symmetric, ε = ∇u.
c) (5 points) Compute the principal strains and principal directions of strain
associated with ε. Give a condition on κ1 , κ2 such that the largest
principal strain is less then 1%.
The principal strains and principal directions of strain are the eigenvalues and eigenvectors of ε, respectively.
2
Since ε is diagonal, the diagonal entries gives the eigenvalues. Then
the principal strains are λ1 = κ1 , λ2 = κ2 and λ3 = 0.
Without loss of generality, one may assume that that κ1 ≥ κ2 > 0.
Then the largest principal strain is κ1 and for the largest principal
strain to be less then 1%, we must have that κ1 < 0.01.
d) (5 points) Compute the Cauchy stress tensor of the body, assuming that
the body is isotropic, homogeneous and linearly elastic with Lame parameters µ = 2.0 MPa, λ = 100.0 MPa.
Under these assumptions, isotropic Hooke’s law holds, and Cauchy’s
stress tensor is given by
(10)
σ = 2µε + λ tr εI
Then σij = 0 for all i 6= j while
(11)
(12)
(13)
σ11 = 2µκ1 + λ(κ1 + κ2 )
σ22 = 2µκ2 + λ(κ1 + κ2 )
σ33 = λ(κ1 + κ2 )
Inserting µ = 2.0 and λ = 100.0 gives the answer.
Oppgave 3
Assume a linear regime with small strains and no distinction between Eulerian and Lagrangian coordinates.
Consider a two-dimensional rectangular body of length a (m) and height
b (m) with coordinates (x1 , x2 ) ∈ [0, a] × [0, b]. Assume that the body is
isotropic and homogeneous with Lamé parameters µ and λ and density ρ.
We shall consider the case where the body is clamped at the ends where
x1 = 0 or x1 = a (and x2 ∈ [0, b]), hence u = (0, 0) there.
Assume that a constant body force f acts downwards on the beam, that
is: f = (0, −g) for some constant g (N/m2 )).
a) (5 points) State the dynamic elasticity equations describing the displacement u = u(x, t) = (u1 (x, t), u2 (x, t)) of a homogeneous, isotropic,
linearly elastic body with coordinates x ∈ R2 , t ∈ (0, T ), Lamé parameters µ and λ and density ρ and a given body force f .
The dynamic elasticity equations read:
(14)
(15)
ρü − ∇ · σ = f,
σ = 2µε + λ tr εI,
(16)
2ε = ∇u + ∇uT ,
3
b) (5 points) Assume that the body is in elastic equilibrium and that there
are no normal stresses (defined relative to the planes orthogonal to the
Cartesian axes) ). There may be tangential stresses on all boundaries.
Compute the resulting stress tensor σ.
If the body is in elastic equilibrium ü = 0, and so balance of momentum
reduces to
−∇ · σ = f,
(17)
By the assumption of no normal stresses and the symmetry of the stress
tensor, we have that
0 σ12
(18)
σ=
σ12 0
Then
(19)
∇ · σ(x1 , x2 ) = (
∂
∂
σ12 (x1 , x2 ),
σ12 (x1 , x2 )) = (0, g)
∂x2
∂x1
Integrating, we find that
(20)
σ12 (x1 , x2 ) = gx1 + c(x2 ),
for some function c of x2 .
c) (5 points) From the result in b), compute the corresponding strain tensor
in terms of the applied body force f and the material parameters µ
and λ.
The strain tensor is given by the inverse isotropic Hooke’s law. In
particular, ε11 = ε22 = 0, while by symmetry and since the trace of σ
vanishes:
(21)
ε12 = ε21 =
1
1
σ12 =
(gx1 + c(x2 ))
2µ
2µ
d) (5 points) From the result in c), compute the corresponding displacement
in terms of the applied body force f and the material parameters µ and
λ.
Since there are no diagonal strains,
(22)
∂
∂
u1 (x1 , x2 ) =
u2 (x1 , x2 ) = 0.
∂x1
∂x2
4
Thus (with a modest abuse of notation)
(23)
u1 (x1 , x2 ) = u1 (x2 ),
u2 (x1 , x2 ) = u2 (x1 ).
Since the body is clamped (u = (0, 0)) for all x2 when x1 = 0, we must
have that
(24)
u1 (x2 ) = 0
By definition of the linear strain tensor and the above, we must have
that
(25)
2ε12 =
∂
1
u2 (x1 ) = (gx1 + c(x2 ))
∂x1
µ
Thus c(x2 ) = C for some constant C. Integrating with respect to x1
we find that
(26)
u2 (x1 ) =
1 g
( x1 (x1 + C) + B)
µ 2
for some constant B. Since u2 = 0 at x1 = 0, B = 0. Since u2 = 0 at
x1 = a, C = −a. Thus, the displacement is
(27)
u(x1 , x2 ) = (0,
g
x1 (x1 − a))
2µ
Hint: Recall the inverse (isotropic, homogeneous) Hooke’s law
1
λ
tr σI
(28)
ε=
σ−
2µ
3λ + 2µ
5