# Transformation relations tensors

1. Aug 24, 2013

### roldy

I'm trying to understand the transformation relations for 2d stress and the book doesn't show the derivation of the 2d stress transformation relations from the directional cosines. The 2d stress transformation relations are found by using the transformation equation and the 2d directional cosines matrix. I'm really confused as to how they go about performing the math.

2d stress transformation relations:

$\sigma_{xx}^{'}= \sigma_{xx} \cos^2 \theta + \sigma_{yy} \sin^2 \theta + 2\sigma_{xy} \cos \theta \sin \theta$

$\sigma_{yy}^{'}= \sigma_{xx} \sin^2 \theta + \sigma_{yy} \cos^2 \theta - 2\sigma_{xy} \cos \theta \sin \theta$

$\sigma_{xy}^{'}= \sigma_{xx}(\cos^2 \theta - sin^2 \theta) + (\sigma_{yy} - \sigma_{xx}) \sin \theta \cos \theta$

transformation equation:

$\sigma_{ij}^{'} = m_{ip} m_{jp} \sigma_{pq}$

2d directional cosine matrix:

$m_{ij} = \left[\stackrel{\cos \theta}{ -\sin \theta}\ \stackrel{\sin \theta}{\cos \theta} \right]$

I guess the thing that I'm confused about is $\sigma_{pq}$. What does that matrix look like?

2. Aug 24, 2013

### mathman

I suspect σpq is simply the original matrix elements.
σ11 = σxx etc.

$\sigma_{ij}^{'} = m_{ip} m_{jp} \sigma_{pq}$
should be
$\sigma_{ij}^{'} = m_{ip} m_{jq} \sigma_{pq}$