Polymers- Plastic Deformation

Question 1

Nominal stress

Answer 1

Force over original area

Question 2

Nominal stress vs strain graph

Answer 2

Linear for strains below 0.001. Reaches a rounded peak (yield point) and then goes back down a bit (strain 0.001-0.01). Then horizontal (dσ/dε is 0) for strains up to 0.1. Finally curves up to failure

Question 3

Temperature dependence of stress strain behaviour

Answer 3

Lower temperature has steeper linear region and yield point at same strain. So lower temperature has higher overall curve

Question 4

Constant volume condition

Answer 4

AiLi=AL
A is CSA, L is length
Means Ai=Aλ where λ is extension ratio L/Li

Question 5

True stress

Answer 5

σt=F/A=λF/Ai=λ σn

Question 6

Relations at the yield point

Answer 6

dσn/dεn=dσn/dλ=(1/λ)dσt/dλ - σt/λ^2
But at yield dσn/dεn=0
So
dσt/dλ=σt/λ

Question 7

Eyring’s model and assumptions

Answer 7

There is an initial state (above some ground level) and then a jump of ΔH as a shear stress is applied and then comes back down to a final state (below the ground level). Like bell curve.
Assumes impose strain rate is proportional to the net jump rate in the direction of the shear stress and the dominant shear stress in a tensile test is the maximum shear stress and at the yield point σs=σy/2

Question 8

Jump rates in Eyring’s model

Answer 8

Before shear stress applied jump rate=αexp(-ΔH/RT)
After applied forward jump rate=αexp(-(ΔH-σsV)/RT)
Net jump rate = αexp(-ΔH/RT)exp(σsV/RT)

Question 9

Imposed strain rate in Eyring’s model

Answer 9

At yield point ε•=dε/dt is:

εy•=ε0•exp(-ΔH/RT)exp(σyV*/2RT)

Question 10

Formula for yield stress in Eyring’s model

Answer 10

σy/w=(2/V*)(ΔH/T + 2.303Rlog(εy•/ε0•)

So yield strength over T is linear increase with log(strain rate) (line lower for greater temperatures

Question 11

Tresca yield criterion

Answer 11

Yield occurs when a critical value of maximum shear stress (σs) is reached.
(σ1-σ3)2=σs
Where σ1>σ2>σ3

Question 12

Von Mises yield criterion

Answer 12

Yield occurs when the shear-strain energy in the material reaches a critical value.
(σ1-σ2)^2+(σ2-σ3)^2+(σ3-σ1)^2=2σy^2

Question 13

Graph of σ1/σy vs σ2/σy for Tresca and Von Mises

Answer 13

Tresca has squares in first and third quadrants with diagonal lines between the corners through the other two quadrants. Von Mises forms an oval that goes through all the outer corners of Tresca. Inside the shapes on the graph is elastic deformation and outside is plastic

Question 14

Where does a craze form?

Answer 14

On a free surface of the object with its length perpendicular to the tensile stress applied

Question 15

Mechanism of crazing

Answer 15

A local-yielding and cold-drawing process which takes place in a constrained zone of a material. Local stresses at the tip of a surface flaw or a growing craze are higher than the overall tensile stress applied to the specimen. Yielding takes place locally at these stress concentrations. Stress required is below that to cause general yielding in the rest of the specimen.

Question 16

Formula for maximum tensile strain

Answer 16

ε1=1/E(σ1-ν σ2-ν σ3)

ν is poisson’s ration

Question 17

When does crazing occur?

Answer 17

At the point when ε1=εc
Accounting for hydrostatic tension εc=C+D/p
σ1 - ν σ2 - ν σ3=X+Y/(σ1+σ2+σ3)
X and H are time and temperature dependent constants

Question 18

Graph of σ1 vs σ2 for crazing

Answer 18

Same sir if shape as Von Mises but top right is cut away as in p13 week 10 and crazing happens beyond this new outline. Shear yielding occurs elsewhere on the unmodified outline

Question 19

Why is crazing more likely for tensile stresses?

Answer 19

Because crazing increases the volume of the specimen and compression would act against this effect