Micromechanics- Classical Laminate Theory 2

Question 1

How are the 9 stress components acting on a small element of material oriented and named?

Answer 1

Consider a cube in standard view. The stress tensor is of the form σij with i and j subscript. For each component the stress acts on a plane normal to the i-axis and acts in the j direction. The planes are the faces of the cube. When i=j the stress is normal σ, if not then the stress is shear τ. The directions and axes are either 1, 2, 3. In the example 1 is up, 2 is towards you and 3 is right.

Question 2

What happens to the stress t’endors when the body is at equilibrium?

Answer 2

σ12=σ21
σ13=σ31
σ23=σ32

Question 3

In tensor notation what are the 3 normal stresses and 3 shear stresses?

Answer 3

Normal: σ11, σ22, σ33
Shear: τ12, τ13, τ23

Question 4

How does contracted notation work?

Answer 4

For i=j subscript is written once so σ11=σ1
For i not j subscript is 9-i-j so σ12=σ6, σ13=5

Question 5

What is different about tensor notation for strain tensor?

Answer 5

For shear strain ε12=γ12/2
The contracted notation has ε6/2 etc

Question 6

Difference between simple shear and pure shear

Answer 6

Simple has on side of square element fixed with opposite side under shear stress τ. Relation is τij =G γij where γ is the engineering shear strain (angle).
Pure shear has all sides under shear stress. Stress produces an average tensor shear strain εij with the relation εij =γij/2. There are two angles of εij which are half as big as γij in simple shear

Question 7

Contracted notation of shear strain

Answer 7

εk=2εij=γij

Question 8

How to transform a stress to a new coordinate frame

Answer 8

Transform stress σ’kl so
σij=a(ik)a(jl)σ’kl
Where aik is the direction cosine of direction i referred to direction k.
ajl is the direction cosine of direction j referred to direction l.
Basically just trigonometry using dot product and vectors
Works the same with strain as well. ‘ means applied

Question 9

Hooke’s law with tensor notation

Answer 9

σij=Cijkl x εkl
So stress field tensor and strain field tensor (both 2nd ranked tensor) are related by the elasticity or stiffness matter tensor C which is a 4th ranked tensor. C is stiffness tensor or stiffness matrix

Question 10

What does each quadrant of the stiffness matrix show?

Answer 10

Top left normal normal
Bottom left shear normal
Top right normal shear
Bottom right shear shear

Question 11

What is the compliance tensor/matrix?

Answer 11

Symbol S. Relates strain matrix to stress matrix by being multiplied by stress matrix (1x6)

Question 12

How many distinct coefficients are required for isotropic materials?

Answer 12

Only 2. S11 and S12 ni contracted notation
S11=1/E
S12=-ν/E

Question 13

Are long fibre composites isotropic?

Answer 13

No but often orthotropic (3 mutually perpendicular planes of symmetry) and transversely isotropic (2 equivalent directions so could rotate composite and get same results)

Question 14

What does it mean when there is no extension-shear coupling?

Answer 14

No interaction between normal stresses and shear strains or shear stresses and normal strains

Question 15

For a lamina loaded in plane stress, what happens to some of the stress tensors?

Answer 15

σ3=σ4=σ5=0

Question 16

For uniaxially applied stress on a lamina what is needed to find the compliance matrix?

Answer 16

E1, E2, ν12, G12

Question 17

What is a lamina loaded along the principal material axes referred to as?

Answer 17

Specially orthotropic. Loaded in 0 and 90° directions only

Question 18

Working with the compliance matrix for specially orthotropic case

Answer 18

See slide 21/22 lecture 4

Question 19

What happens when a lamina is loaded at an arbitrary angle?

Answer 19

The lamina is ‘generally orthotropic’. Will be normal-shear interactions Si number of non-zero compliance coefficients rises

Question 20

Transformation matrix

Answer 20

In the databook. Used to express the applied stresses [σ’] in terms of stresses in the principal material axes [σ]. Multiplied by the σ’ matrix. c means cosθ and s means sinθ. Where θ is angle between applied and principal directions. Can also be used for tensorial strains but not engineering strains

Question 21

Which transformation matrix is needed for engineering strains?

Answer 21

For when γ12=2ε12. Need transformation matrix U in the databook. U multiplied by ε’ matrix

Question 22

Transformed compliance tensor

Answer 22

[S bar] used for
[ε’]=[S bar][σ’]
This for relating applied stress to applied strain at any angle
Equal to [U]^-1 [S][T]

Question 23

Individual elements of transformed compliance tensor

Answer 23

Formula for each in databook for generally orthotropic or general case

Question 24

Composite YM in longitudinal and transverse direction against loading angle

Answer 24

E’1 and E’2. E’1 starts high at E1 then curves down with soon inflexion then settles down at E2 up to 90°. E’2 is opposite shape starting at low E2 going up to high E1

Question 25

Composite shear modulus vs loading angle

Answer 25

Like a bell curve centred at 45°. Lowest points at 0 and 90 are G12

Question 26

Major and minor Poisson’s ratio vs loading angle

Answer 26

ν’12 starts medium at ν12 then up then curve down to ν21 at 90°. Opposite shape for ν’21 starting low at ν21 then going high and settling at medium ν12

Question 27

Composite interaction ratios

Answer 27

η’121=E1Sbar16
η’122=E2Sbar26
First one is ratio of shear strain caused by normal stress to normal strain caused by normal stress
Higher magnitude values mean more shear-normal interaction occurs

Question 28

Composite interaction ratios vs loading angle

Answer 28

η’122 starts 0 then up a bit then curve down lots then steeply back up to 0 at 90°.
η’121 opposite shape but still starts and ends at 0