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Question 1

Describe the essential features of covalent bonding and van der Waals interactions in polymers.

Answer 1

Covalent Bonding: The primary force holding polymers together, involving the sharing of electron pairs between atoms. In polymers, covalent bonds form the backbone of the polymer chain, providing strength and stability.
Van der Waals Interactions: Weaker than covalent bonds, these are intermolecular forces that include attractions and repulsions between atoms, molecules, and surfaces. In polymers, they affect the transition temperature, mechanical properties, and solubility.

Question 2

. Compare the structure and hierarchy between polymers and metals. Describe terms that may have
different names but similar meaning.

Answer 2

Polymers: Typically have a hierarchical structure that includes the primary covalent chain, secondary van der Waals or hydrogen bonding, tertiary folding or entanglement, and quaternary cross-linking or network structure.
Metals: Characterized by a crystalline structure with atoms arranged in a regular, repeating pattern. Metallic bonding provides a sea of delocalized electrons contributing to properties like conductivity and ductility.
Comparison: While polymers may have long-range order in the case of crystalline regions, they are generally more disordered than metals. Terms like grain or phase can refer to regions of different structural order in both classes of materials.

Question 3

Draw representative polymers depicting 1D and 3D structures.

Answer 3

1D Structure: Typically represents the linear chain of a polymer, showcasing the repeat units and covalent bonds.
3D Structure: Shows how polymer chains can be arranged in space, including branching, cross-linking, and folding, contributing to the macroscopic properties.

Question 4

Describe how structural properties of a polymer may alter the mechanical properties.

Answer 4

The structural properties such as molecular weight, chain branching, and cross-linking significantly influence mechanical properties. For example, higher molecular weight or cross-linking density can lead to increased tensile strength and modulus.

Question 5

Polymers are known to have ’time-dependent’ properties. With respect to viscoelasticity, explain how
time plays a role in polymer mechanics. With respect to biodegradability, explain how time plays a
role in polymer mechanics.

Answer 5

Viscoelasticity: Refers to the time-dependent deformation of polymers, where they exhibit both elastic and viscous responses to applied stress or strain. Over time, polymers may show creep (increased deformation under constant load) or stress relaxation (reduced stress under constant deformation).
Biodegradability: The degradation rate of polymers depends on environmental conditions and the chemical structure of the polymer. Time plays a critical role in the biodegradation process, as it dictates the duration for which the polymer remains intact in the environment.

Question 6

In your own words explain cradle-to-grave/green design.

Answer 6

This concept involves considering the environmental impact of a product from its creation (cradle) to disposal (grave). In green design, the aim is to minimize the negative environmental impact through sustainable design practices, including the use of renewable resources, energy efficiency, and recyclability.

Question 7

Describe the relationship between polymer structure and mechanical behavior in uniaxial tension.

Answer 7

The mechanical behavior of polymers under uniaxial tension is influenced by their molecular structure. For example, highly crystalline polymers exhibit higher strength and stiffness, while amorphous regions contribute to ductility and toughness.

Question 8

Identify and describe polymer defects that are similar to defects discussed for metals.

Answer 8

Dislocations in crystalline polymers can be analogous to those in metals, affecting mechanical properties like yield strength and ductility.
Void formation and inclusions can act as stress concentrators, leading to premature failure, similar to metals

Question 9

Identify and describe polymer defects that are unique to the polymer chain structure.

Answer 9

Chain entanglements can act as physical cross-links, affecting the viscosity and mechanical strength.
Unreacted monomers, oligomers, or branching can create weak points in the material, affecting its overall properties.

Question 10

Describe at least three methods for manufacturing or processing polymers.

Answer 10

Extrusion: Forces melted polymer through a die to form shapes like tubes, sheets, or profiles.
Injection Molding: Molten polymer is injected into a mold, where it cools and solidifies into the desired shape.
Blow Molding: Used for making hollow objects, such as bottles, by blowing air into a heated polymer tube, which expands into a mold.

Question 11

What is a rheological model and how does it differ from a constitutive model?

Answer 11

Rheological Model: Focuses on the flow and deformation behavior of materials, especially under applied stresses, and is primarily used for fluids and viscoelastic materials like polymers.
Constitutive Model: A more general term that describes the stress-strain relationship of materials under various loading conditions. It encompasses rheological models as a subset, often used for both solid and fluid mechanics.

Question 12

Describe the three rheological models presented in class.

Answer 12

Maxwell Model: Consists of a spring (elastic element) and a dashpot (viscous element) in series. It represents materials that exhibit immediate elastic deformation followed by viscous flow.
Voigt Model: Features a spring and dashpot in parallel, representing materials that exhibit delayed elastic deformation.
Standard Linear Solid (SLS) Model: Combines the elements of both Maxwell and Voigt models, showing both immediate elastic response and time-dependent viscous effects.

Question 13

Be able to derive a simple rheological model. In class we worked through the derivation for the
Maxwell model in creep and stress-relaxation. Work through the derivation for the Voigt model.

Answer 13

In the Voigt model, the total strain is the sum of strains in the spring and dashpot, which are in parallel. If a constant stress
σ is applied:
σ=Eϵe=η dt/dϵ

Where
E is the elastic modulus of the spring,
η is the viscosity of the dashpot,
ϵ e is the elastic strain, and
ϵ v is the viscous strain.
To derive the model, integrate the equation to express strain as a function of time, showing the delayed elastic response characteristic of the Voigt model.

Question 14

Draw the creep and stress-relaxation curves for the (a) Maxwell, (b) Voigt, and (c) SLS models.

Answer 14

(a) Maxwell Model:
Creep: Shows an immediate elastic deformation followed by a constant rate of viscous flow.
Stress-Relaxation: Rapid decrease in stress, reflecting immediate elastic deformation followed by stress relaxation due to viscous flow.
(b) Voigt Model:
Creep: Exhibits a gradual increase in strain over time, reaching a steady state.
Stress-Relaxation: Stress remains constant over time, as the model cannot exhibit stress relaxation due to the parallel arrangement of elements.
(c) SLS Model:
Creep: Initial rapid strain increase (like Maxwell), followed by a slower, steady increase (like Voigt).
Stress-Relaxation: Shows a rapid initial stress drop (like Maxwell) followed by a slower decline (like Voigt).

Question 15

For a given creep or stress-relaxation response draw the viscous and elastic responses separately.

Answer 15

For each model, you can depict the viscous and elastic responses separately by considering the behavior of the dashpot and spring elements independently.
In the Maxwell model, the elastic response is immediate deformation, while the viscous response is a linear increase in strain over time.
In the Voigt model, the elastic response is delayed, and the viscous response contributes to the gradual, continuous deformation.
In the SLS model, the responses are combined; the elastic response includes both immediate and delayed deformation, while the viscous response adds a time-dependent rate of strain increase.

Question 16

1. Derive stiffness matrix for a stress component using Hooke’s Law.

Answer 16

google

Question 17

Draw planar stress state including normal and shear stresses.

Answer 17

google

Question 18

Draw a complex stress state on a 3D engineering component, including all 6 potential stress
conditions.

Answer 18

google

Question 19

. Given a complex stress state, draw a Mohr’s circle and solve for:
average stress,
radius of the Mohr’s circle,
maximum principal stress,
minimum principal stress, and
rotation angle needed to convert the complex stress state to a condition maximum normal
stresses (and no shear) or maximum shear stress.

Answer 19

google

Question 20

Be able to describe when the maximum normal stress criterion would be used rather than the
maximum shear stress criterion for failure.

Answer 20

The maximum normal stress criterion is used when failure occurs due to normal stress, as in brittle materials like ceramics and some high-strength steels. It’s applicable when the material fails upon reaching its tensile or compressive strength limit before yielding.

Question 21

. Provide examples of materials where 1) maximum normal stress or 2) maximum shear stress criterion
would be a better approach.

Answer 21

Maximum Normal Stress Criterion: Brittle materials, such as ceramics and glass, where failure is due to fracturing under normal stress.
Maximum Shear Stress Criterion (Tresca): Ductile materials, like mild steel and aluminum, where yielding precedes fracture and failure occurs due to shear deformation.

Question 22

. Consider the low-alloy stress-strain response shown below for low alloy steels. As the carbon content
increases, strength increases at the loss of ductility.
Of the two failure criterion discussed in class today, which one would you used for AISI 1065 or AISI
1095 versus AISI 1020? Be able to support your answer.

Answer 22

For AISI 1065 and 1095 (high carbon content, higher strength, less ductility), the maximum normal stress criterion might be more appropriate because these steels are likely to fail at lower strains under high normal stresses.
For AISI 1020 (low carbon content, less strength, more ductility), the maximum shear stress criterion (Tresca) would be suitable as it better predicts the yielding behavior which occurs before fracture.

Question 23

. If you only knew σy and σu,tension but did not know what the stress-strain response looked like,
which criterion would you use, and why?

Answer 23

Without knowing the stress-strain response, the choice depends on the material type (ductile or brittle). Generally, if
σy is significantly lower than σu,tension
, indicating ductility, Tresca (shear stress criterion) may be preferable. For materials where σ y and ,tension σu,tension
​are close, indicating brittleness, the maximum normal stress criterion might be used.

Question 24

. Be able to calculate the factor of safety for an engineering element based on the design and use the
information to predict whether the part would experience failure.

Answer 24

The factor of safety (FoS) is calculated as
FoS
=σ actual/σ allowable
, where allowable σ allowable
is the material’s yield strength or ultimate strength divided by the factor of safety, and
actual σ actual
is the stress experienced by the material.
To predict failure, compare the FoS with 1. If FoS > 1, the part is likely safe; if FoS < 1, the part may fail under the given load.
By understanding these criteria and factors, engineer

Question 25

User
Explain what mechanical properties are measured during a Charpy or Izod test. Be able to describe
the physics behind the test and how it realates to the stress-strain curve.

Answer 25

Measured Properties: These tests measure the impact energy, i.e., the energy absorbed by a material during fracture when a sudden impact or force is applied. The result, expressed in Joules, indicates the material’s toughness – its ability to absorb energy and plastically deform before fracturing.
Physics Behind the Test: In both tests, a pendulum hammer is swung to hit a notched specimen, breaking it at the notch. The energy absorbed by the specimen to fracture is calculated based on the height to which the pendulum swings after hitting the specimen. This energy absorption is directly related to the area under the material’s stress-strain curve, representing the toughness.
Relation to Stress-Strain Curve: The total area under the stress-strain curve is a measure of the material’s toughness. In these tests, the energy absorbed (toughness) corresponds to the ability of the material to withstand impact loads without fracturing, reflecting both the elastic and plastic deformation regions of the stress-strain curve.

Question 26

In your own words describe the impact of cracks on material behavior.

Answer 26

Cracks act as stress concentrators in a material, significantly increasing the local stress near the crack tip. This elevated stress can lead to premature material failure at lower applied loads than expected for a crack-free material. The presence of cracks reduces the effective load-bearing area and can significantly alter the material’s mechanical properties, such as strength and toughness.

Question 27

What are microscopic differences between materials at the crack tip edge?

Answer 27

At the microscopic level, materials may show different behaviors at the crack tip, such as plastic deformation in ductile materials or brittle fracture in brittle materials. In ductile materials, the crack tip might undergo significant plastic deformation, leading to a blunted crack tip. In contrast, in brittle materials, the crack tip might propagate rapidly with little to no plastic deformation, leading to sharp crack propagation paths.

Question 28

Describe design methods to reduce stress concentrations.

Answer 28

Avoid Sharp Corners: Use fillets and rounded corners instead of sharp edges to reduce stress concentrations.
Material Selection: Choose materials with higher toughness to withstand stress concentrations better.
Crack Stoppers: Design features that impede crack propagation, such as crack stopper holes or notches strategically placed to interrupt the crack path.
Uniform Section Transitions: Ensure that changes in cross-sectional areas are gradual rather than abrupt to minimize stress concentration.
Proper Loading: Design components to ensure loads are applied evenly and avoid unexpected load paths that may introduce stress concentrators.
Post-Processing Treatments: Apply treatments like heat treatments or surface finishes that can reduce the likelihood of crack initiation or propagation.