Polymer Mechanical Properties

Introduction

A polymer can be brittle, ductile or rubbery depending on chemistry, molar mass, crystallinity, temperature and loading speed. Stress–strain curves reveal how it deforms, while viscoelasticity explains why the same sample responds differently to a quick pull and a long sustained load.

Core explanation

Engineering stress is force divided by original cross-sectional area, and engineering strain is extension divided by original length. The initial slope of a stress–strain curve gives an elastic modulus over a sufficiently small linear region. A glassy polymer may show a steep slope and fracture at modest strain. A ductile thermoplastic may yield, neck, draw and then strengthen as chains align. A cross-linked elastomer above its T g can stretch greatly and recover because the network prevents permanent chain separation. Rubber elasticity is largely entropic: stretching uncoils many chains and reduces their number of available conformations; release lets them return toward more probable coiled states. Polymer response is viscoelastic, combining recoverable elastic deformation and time-dependent flow or relaxation. Under constant stress, strain can increase over time, called creep. Under constant strain, stress can decrease, called stress relaxation. Loading rate and temperature are therefore essential test conditions. Crystallites can serve as physical reinforcing regions, while chain entanglements and high molar mass improve resistance to pull-out. Too much cross-linking can raise stiffness but reduce extensibility. Report the full material state and test conditions rather than interpreting a single modulus as a permanent intrinsic property independent of history.

Step-by-step reasoning

Define the loading geometry, calculate stress F/A₀ and strain ΔL/L₀, and inspect the initial slope for modulus. Identify yield or fracture on the curve. For time-dependent tests, state whether force or deformation is held fixed and predict creep or relaxation. Compare test temperature with T g and any melting transition.

Visual explanation

Draw three curves on stress–strain axes: steep short glassy fracture, a ductile yield-and-draw curve, and a long low-slope elastomer curve. Beside them sketch strain rising under constant stress and stress falling under constant strain.

Real-world analogy

A rubber band stores a tendency to recoil when stretched, while warm putty slowly changes shape under a persistent load. Polymers often show both kinds of response in different proportions depending on timescale and temperature.

Real-world example

Tyre rubber contains a cross-linked network designed to deform and recover repeatedly, while a drawn polymer fibre gains strength from aligned chains. Both are polymers, yet their chain architecture and loading response differ markedly.

Why?

Chain segments move on finite timescales. Quick loading can outpace relaxation and appear stiff; slower loading allows rearrangement. In a rubber network, stretching reduces conformational entropy, providing a restoring force when released.

Common misconception

Elastic modulus is not simply the maximum stress a sample can withstand. It is a slope in the small-strain region. Creep and stress relaxation are also distinct: one measures changing strain at fixed stress, the other changing stress at fixed strain.

Worked example

Question: A specimen initially 50 mm long extends by 2.5 mm under a force of 100 N. Its original area is 10 mm². Find engineering stress and strain. Reasoning: Stress = 100/10 = 10 N mm⁻² = 10 MPa; strain = 2.5/50 = 0.050. Answer: 10 MPa stress and 5.0% strain.

Quick check

1. What is creep in a polymer? Answer: Increasing deformation with time under an approximately constant applied stress.

Exam focus

Label axes and units. Give the test temperature and loading rate when comparing materials, and distinguish modulus, yield stress, tensile strength and strain at break.

Advanced insight

Time–temperature superposition can sometimes relate a polymer's response at short times and high temperatures to response at longer times and lower temperatures. It is useful only when the same relaxation mechanisms remain active across the compared conditions.

Summary

Stress is force per area and strain is relative extension. Glassy, ductile and rubbery polymers show different stress–strain shapes. Viscoelasticity creates creep and stress relaxation, while cross-linked rubber recovers mainly through entropic chain recoil. Temperature and timescale must accompany property values.

Practice questions

1. Calculate engineering strain for a 100 mm sample extended to 105 mm. Answer: (105 − 100)/100 = 0.05, or 5%.

2. Define stress relaxation. Answer: A decrease in stress over time while the imposed strain is held approximately constant.

3. Why does cross-linked rubber recover after stretching? Answer: Network links prevent permanent separation, and chains tend to return to more numerous coiled conformations.

4. Why can a fast tensile test give a different modulus from a slow one? Answer: Segmental relaxation has less time to occur during fast loading.