GENG2004 - Lecture 2

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GENG2004 - Lecture 1
Stress in 3D - Slides

The Stress Tensor

We know that stress is force per unit area that acts upon a surface:
$\sigma =\frac{F}{A}$
Remember stress is measured in Pascals ($Pa$)

However, inside a real material, there are infinitely many possible planes passing through a point → each plane experiences different forces

• ‘Stress’ vectors are in reality known as traction vectors

  
  This vector captures the normal stresses (perpendicular to the surface) and the shear stresses which are non-perpendicular
  ${\lim }_{\Delta S\to 0}\frac{\Delta \text{F}}{\Delta S}=\frac{d\text{F}}{dS}=t_n$

• A traction vector simply is a representation of stress acting upon a particular surface → two components to account for normal + shear

  • i.e. traction vector → total stress acting upon that face

For each surface orientation, you would get a different traction vector

• We use a 3x3 matrix, known as a stress tensor, to store the information to generate these traction vectors for infinitely many planes
  • Each face experiences normal stresses and two shear stresses:
    Normal stress → outward from surface
    Shear stresses → along surface and perpendicular to each other

Stress tensor → gives $\frac{F}{A}$ for any surface direction!

You are expected to know the direction of each stress component, the diagram above has the correct labels used for this unit

${\sigma }_{ij}$ stress tensor notation for this unit:

• $i$ $-$ direction of outer unit normal
• $j$ $-$ direction of traction

In this unit, stress (and strain) are symmetrical, this means:
${\sigma }_{12}={\sigma }_{21},{\sigma }_{13}={\sigma }_{31},{\sigma }_{23}={\sigma }_{32}$
Essentially, since the stress tensor represents internal forces acting upon an infinitesimally small cube in the material, this cube cannot spin independently otherwise the solid would ‘crumble’ and behave like a liquid. We need to equal out the ‘spin’ as we assume the solid is a continuum.

Internal Forces in Solids

• Our traction vector gives us force per unit area on a surface
  → To get force, we multiply traction by a small area:
  $d\text{F}=\text{t}dA$
  When you cut a beam cross-section, internal stresses act across area
• We end up with:
  • Normal force $N$ $-$ from normal stress
  • Shear stresses $V$ $-$ from shear stress
  • Bending moment $M$ $-$ from distribution of normal stress
  • Torsion $T$ $-$ from distribution of shear stress
• We can calculate these values by integrating stress

Normal force:
$N={\int }_A\sigma dA$

Shear force:
$V={\int }_A\tau dA$
shear stress = $\tau$

Bending moment:
$M={\int }_{A}y\cdot \sigma dA$
$y$ is distance from the neutral axis
For $M_y$ you use $y$, but for $M_x$ and $M_z$, use $x$ and $z$ respectively

• Bending happens because normal stresses vary across the cross-section

Torsional moment:
$T={\int }_{A}r\cdot \tau dA$
$r$ is the distance from centre

• Each shear stress creates small force → each creates a twisting moment

Stress Transformations

• Lets say our coordinate system changes:
  • Material stress state → remains the same
  • Numbers in tensor describing it → now change
• We can transform a stress tensor to work in the new coordinate system
  For this example, we will be using $x_1,x_2,x_3$

If the coordinate system rotates, the stress tensor transforms as:
${\sigma }^{\prime }=Q\sigma Q^T$
$\sigma$ → original stress tensor, ${\sigma }^{\prime }$ → stress tensor in rotated axes
$Q$ → rotation matrix, $Q^T$ → transpose of rotation matrix
Transpose → swap rows and columns

For a rotation, we can find $Q$ by solving this matrix:
$Q=\left[\begin{array}{ccc}\cos (x_1^{\prime },x_1)& \cos (x_1^{\prime },x_2)& \cos (x_1^{\prime },x_3)\\ \cos (x_2^{\prime },x_1)& \cos (x_2^{\prime },x_2)& \cos (x_2^{\prime },x_3)\\ \cos (x_3^{\prime },x_1)& \cos (x_3^{\prime },x_2)& \cos (x_3^{\prime },x_3)\end{array}\right]$
$\cos (x_i^{\prime },x_j)$ → cosine of the angle between new axis $x_i^{\prime }$ and old axis $x_j$
e.g. for a 45$°$ rotation about $x_3$, the angle between $x_1^{\prime }{x}_1$ is 45$°$ which is also true for $x_2^{\prime }{x}_2$. For $x_3^{\prime }{x}_3$, the angle remains at 0$°$.

• The angles you are going to get are going to be multiples of the rotation (in this case 45$°$) and you can calculate it in your head

See: Matrix Multiplication

• We multiple each row of left matrix by each column of right matrix to do matrix multiplication

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GENG2004 - Lecture 3