Question

PET has a Young's, Modulus of 3.5 GPa and PLGA 1.2 GPa. A polymer blend is prepared with 40:60 mixture (PET:PlGA) . What is the Young's, Modulus of this blend (use mixture rule). Calculate the percent load borne by PET when the blend is subjected to a load. If after 10 days, half of PET and one third of PLGA had degraded, what is the Young's Modulus of this blend.

Answer 1

Explanations & Calculations

• When mixed, the property of elastic modulus varies according to the mixing ratio.
• Then that is for the blend will be,

$\qquad\qquad \begin{aligned} \small Y_{blend}&=\small 3.5\,GPa\times \frac{40}{100}+1.2\,GPa\times \frac{60}{100}\\ &=\small \bold{2.12\,GPa} \end{aligned}$

$\qquad\qquad ...............................................................$

• When a load is applied on the material it behaves according to the properties of each component materials. A rough estimate of the percent load carried by PET can be obtained by comparing the moduli of the blend & the particular component material.
• Then,

$\qquad\qquad \begin{aligned} \small \frac{F_{PET}}{F_{blend}}\times100&=\small \frac{3.5\times 0.4}{2.12}\times 100=\bold{66.04} \end{aligned}$

$\qquad\qquad...............................................................$

• After the degradation, there exist obsolete material within the blend which causes reduction in properties.
• To calculate the new modulus, the percent amounts of effective material is needed.
• The total volume of the blended material does not change. If the initial volume of PET was $\small V_1$ and that of PIGA was $\small V_2$, then at the beginning

$\qquad\qquad \begin{aligned} \small V_{pet}:V_{piga}&=\small 40:60=2:3 \end{aligned}$

• What is left effective after 10 days is

$\qquad\qquad \begin{aligned} \small V_{pet}&=\frac{2}{5}\times \frac{1}{2}=\frac{1}{5} \\ \small V_{PIGA}&=\small \frac{3}{5}\times \frac{2}{3}=\frac{2}{5} \end{aligned}$

• Then their ratio is

$\qquad\qquad \begin{aligned} \small V_{PET}:V_{PIGA}:V_{obsolete\,material} &=\small 100\times \frac{1}{5}:100\times \frac{2}{5}:100\times\frac{2}{5}\\ &=\small 20:40:40 \end{aligned}$

• Then the new modulus will be

$\qquad\qquad \begin{aligned} \small Y'&=\small 3.5\,GPa\times\frac{20}{100}+2.1\,GPa\times\frac{40}{100}\\ &=\small 0.7+0.84\\ &=\small \bold{1.54 \,GPa} \end{aligned}$