Question 2.19: Figure 2.28 shows the state of stress at a point. Determine ...

Using simpler notations for direction cosines,

let $l_1 , m_1 , n_1$ = direction cosines of $x^{\prime}$ w.r.t. x, y, z axis,

$l_2, m_2, n_2$ = direction cosines of $y^{\prime}$ w.r.t. x, y, z axis,

$l_3, m_3, n_3$ = direction cosines of $z^{\prime}$ w.r.t. x, y, z axis

Since $x^{\prime} ,y^{\prime} and  z^{\prime}$ are mutually perpendicular to each other, their direction cosines will be related as follows.

$l_1 \times l_2 +m_1 \times m_2 + n_1\times n_2=0$ (i)

$l_2 \times l_3 + m_2 \times m_3 + n_2 \times n_3 = 0$ (ii)

$l_3 \times l_1 +m_3\times m_1 +n_3 \times n_1=0$ (iii)

Using Eq. (2.12) for relating direction cosines of $x^{\prime}$-axis,

$(c_{nx})^2+(c_{ny})^2+(c_{nz})^2=1$            (2.12)

$l_1 ^{2} + m_1^{2} + n_1^{2}=1$ (iv)

Since $x^{\prime}$-axis is inclined equally w.r.t. x, y and z directions,
$l_1= m_1= n_1= 1/\sqrt{3} = \pm 0.5773$

Using only +ve value for further calculations, substitute these values in Eq. (i),
$l_2 +m_2+n_2=0$ (v)

But, $l_2$ = cos 60° = 0.5

Substituting in Eq. (2.12),

or, $n_2= 0.309  and  m_2= -0.809$

Similarly using Eqs (ii), (iii) and (2.12),

Stress tensor in rotated $x^{\prime}y^{\prime} z^{\prime}$ reference direction is obtained by using Eq. (2.10d) in matrix notation,

$\tau _{ns}=\sum\limits_{i}\sum\limits_{j}\times \tau_{ij}\times c_{ni}\times c_{nj}$                (2.10d)