
Engineering
227: Advanced Elasticity
Problem Set 3
Due Wednesday, October 22, 2003
- A
Neoprene rubber bar is subjected to a uniaxial tensile test. The bar’s
initial cross sectional area was 1cm2. Initially, the
deformations are kept small and the bar is found to have Young’s modulus
of 4.25 MPa. The extension is continued until the bar breaks, which occurs
when the bar is stretched to 4.2 times its original length. Assume that
the elastic potential has the form described in class: .
- Find
the value of the material modulus m
in MPa.
- Find
the tensile strength of the material (maximum sustainable stress in
tension) in MPa.
- What
was the maximum force applied to the bar during the test? Give the answer
in Newtons and pounds.
- What
was the cross sectional area just before the bar broke?
- Consider
an infinitesimal homogeneous deformation with displacement gradient tensor
b: Assume in what follows that the
reference configuration is stress free.
- Show
that the nominal stress is related to the displacement gradient b through the linear relations
where is called the elasticity four tensor.
- Show
that the nominal and Cauchy stress are related through .
Show further that the balance of angular momentum requires that the
elasticity tensor have the first minor symmetry cijkl= cjikl.
- Suppose
the material is hyperelastic, so that Show further that the elasticity tensor
has major symmetry cijkl=
cklij. If the material is not hyperelastic, is it still
possible for the elasticity tensor to have major symmetry? Why or why
not?
- Assume
that the elasticity tensor does have major symmetry. Use this and the
result of part (b) to show that the elasticity tensor have the second
minor symmetry cijkl=
cijlk.
- Next,
show with the aid of part (d) that
the stress-strain relations can be written as with the error of order e2. The tensor is the infinitesimal strain tensor.
- Show
that
- Now
comes objectivity and material symmetry: recall that for every proper
orthogonal Q and tensor F with positive determinant,
objectivity demands For solids, with the material symmetry
group is a subset of the proper orthogonal group: .
Then, Putting these together gives By differentiating this identity with
respect to F and evaluating the result at F=I, show that this implies for every Q in the material
symmetry group.
- When
subjected to small deformations, a material has a stress-strain law given
by in which are material constants and n is a unit vector.
- Show
that the material is elastic and find the elastic potential
- What
type symmetry does this solid have?
- The
solid is in a state of uniaxial stress parallel to the e1-direction; the
material is oriented such that n=e2. One finds that the
non-zero stress and strain components are related through . Relate to .
Check the results by setting .
- Again,
the solid is in a state of uniaxial stress parallel to the e1-direction. This
time, the material is oriented such that n=e1.
What are the resulting strain components, and how do they relate to ,
and ?
Again, check the results by setting .
- A
nonlinear spring is made from a rubber cylindrical tube (inner radius a, outer radius b, height h) as
shown below. The outside of this tube is cemented to a fixed, rigid
container and the inside to a rigid shaft; the end faces of cylinder are
traction free. In order to induce a net shaft displacement d, a force F is applied to the shaft end. The elastic potential for the
material is the one you already know and love, from Problem Set 2, .


The goal is to find F given d, and sketch a force-displacement
curve for the spring. The boundary value problem consists of satisfying
the usual equations of equilibrium subject to the boundary conditions: on ,
on ,
and on .
Consider an axially symmetric deformation of the form . There are no body forces.
- Show
that the Cartesian components of P in this basis are as
follows: .
Here, ,
and is the material constant.
- Before
you attempt to satisfy the equilibrium equations, see if this stress
field consistent with the traction-free end conditions. If you can’t
satisfy these end conditions exactly at every point on ,
see if it is possible to satisfy the end conditions in an integral sense,
regardless of the actual form taken by u(r). Thus, show
that the tractions generated by any deformation of the above form produce
no net force (in any of the
three directions) on the end-faces, and no net moment (about any of the
three axes).
- Now
use the balance of linear momentum (equilibrium equations) and boundary
conditions to find the function u(r) in terms of ,
and d. This function u(r) gives an approximate solution to the original boundary
value problem.
- Using
the function u(r) determined above, find the net
force F required to displace
the inner bar, and sketch a curve of F
versus d.
- Find
the components of Cauchy stress.