Exam I 2013 - Iowa State University

Mat E 454/MSE 545 Polymer Composites and
Dr. Michael R. Kessler
Exam I - Spring 2013
Problem 1
/7 .
Problem 2
/12 .
Problem 3
/8 .
Problem 4
/16 .
Problem 5
/16 .
Problem 6
/16 .
Problem 7
/10 .
Problem 8
/15 .
Problem 1.
Match each statement on the left with one topic from the box.
Points 7
1. Nucleation and Growth
2. Homogenization
a) This staining agent is often used to improve contrast between
phases for TEM.
3. Spinodal decomposition
4. Solubility Parameter
b) The square root of the cohesive energy density.
5. Inverse rule of mixtures
c) The shape and arrangements of phases.
6. Large-angle X-Ray
d) Useful technique for studying domain sizes in block
copolymers, blends, and nanocomposites from 5-100 nm.
7. Morphology
e) Defines a reinforcement efficiency to obtain a better estimate
than the inverse rule of mixtures.
f) The phase separation kinetics for microstructures found
between the spinodal and bimodal curves.
8. Flory-Huggins
9. Osmium Tetroxide
10. Small-angle X-Ray
11. Halpin-Tsai Equation
g) Polymerization of monomer II in the presence of polymer I.
Problem 2.
12. Reaction induced
phase separation
Points 12
Draw a typical DSC curve for a semi-crystalline polymer and label the glass transition, the
crystallization peak, and the melting peak. Make sure to indicate which direction is exothermic,
and include units on your graph.
Problem 3.
Points 8
True or False
a) Increasing molecular weight increases entropy of mixing
b) As the difference in solubility parameters between two components increases, the FloryHuggins interaction parameter, χ1, decreases.
c) Most polymer blends have a phase behavior exhibiting a lower critical solution temperature.
e) If the Gibbs free energy of mixing, ΔGm, is less than zero, than phase separation will occur.
f) If the Flory-Huggins interaction parameter, χ1, is negative, the Flory Huggins theory will
always yield a negative value of ΔGm.
g) The Stress Intensity Factor, K, is proportional to the stress near a crack tip.
h) Poisson’s ratio for an ideal (incompressible) rubber is 1.
l) Polymer-polymer phase behavior is controlled by the following four factors: (1) choice of
monomer, (2) molecular architecture, (3) composition, and (4) molecular size.
Problem 4.
Points 16
Poly(cy) (Tg=140C) is mixed with Poly(clones) (Tg= –40C) at a 65/35 weight ratio. You
measure the glass transition temperature of the blend and find two distinct glass transitions, one
at –5C and one at 100C.
(a) What is the composition of each phase?
(b) What is the volume fraction of each phase (assume equal densities)?
Problem 5.
Points 16
The Inverse Rule of Mixtures for Transverse
Stiffness is
V f Vm
Ec E f Em
(a) Derive this equation starting with an iso-stress
c   f  m
(b) What are the assumptions in the derivation?
(c) In general are they valid for a unidirectional fiber reinforced polymer composite? Why or
why not?
(d) What alternative micromechanical model for transverse stiffness would you suggest?
Problem 6.
Points 16
You want to create a miscible polymer blend with poly(methyl methacrylate) and poly(vinyl
chloride) at 200ºC with a 20% volume fraction for PMMA and a 80% volume fraction for PVC.
Assuming the PVC has a uniform molecular weight of 60,000 g/mol, what is the maximum
molecular weight of the PMMA?
The atomic mass of chlorine is 35.5,
oxygen is 16, carbon is 12, hydrogen is 1
Assume Vr is obtained by the geometric mean:
𝑉𝑟 = √(𝑉𝑟𝑃𝑀𝑀𝐴 )(𝑉𝑟𝑃𝑉𝐶 )
Problem 7.
Points 10
Estimate the Solubility Parameter for poly(butyl methacrylate) (see structure below).
density of PBMA is 1.053 g/cm3.
Problem 8.
Points 15
a) Sketch and label the three modes of fracture.
b) A new fracture specimen has been designed. The compliance of this specimen as a function
of crack length is found to be
Where C is compliance, E is Young’s modulus, b is specimen thickness, d is a constant equal to
8 m-1, and a is crack length. An isotropic composite material is tested using this specimen
(specimen is thin so that plane stress can be assumed). The fracture toughness for the material,
KIC, is 2.0 MPa√m. What is the critical load for failure (Pc) if the specimen has a crack length of
a=45mm and a specimen thickness b of 1 mm? Comment on the stability of the specimen under
fixed load conditions?
Values of R
8.314472 J · K-1 · mol-1
0.08205746 L · atm · K-1 · mol-1
8.2057459 x 10-5 m³ · atm · K-1 · mol-1
8.314472 L · kPa · K-1 · mol-1
62.3637 L · mmHg · K-1 · mol-1
62.3637 L · Torr · K-1 · mol-1
83.14472 L · mbar · K-1 · mol-1
1.987 cal · K-1 · mol-1
10.7316 ft³ · psi · °R-1 · lbmol-1
NA= 6.0221415×1023 units/mole