Part (a)

From the previous problem, for an angle of rotation of 30${\displaystyle ^{o}}$, the rotation matrix ${\displaystyle \left[L\right]}$ is

${\displaystyle l_{ij}=\left[L\right]={\begin{bmatrix}{\sqrt {3}}/2&1/2&0\\-1/2&{\sqrt {3}}/2&0\\0&0&1\end{bmatrix}}}$

Therefore, the components of strain in the rotated co-ordinate system are given by

${\displaystyle \left[{\boldsymbol {\varepsilon }}\right]^{'}=\left[L\right]\left[{\boldsymbol {\varepsilon }}\right]\left[L\right]^{T}~~{\text{or,}}~~\varepsilon _{ij}^{'}=l_{ip}l_{jq}\varepsilon _{pq}}$

Since we are given ${\displaystyle \varepsilon _{30^{o}}=\varepsilon _{11}^{'}}$, we will calculate the value of this strain in terms of the original components of strain. Thus,

${\displaystyle {\begin{aligned}\varepsilon _{11}^{'}=&l_{1p}l_{1q}\varepsilon _{pq}\\=&l_{11}l_{11}\varepsilon _{11}+l_{12}l_{11}\varepsilon _{21}+l_{13}l_{11}\varepsilon _{31}+l_{11}l_{12}\varepsilon _{12}+l_{12}l_{12}\varepsilon _{22}+l_{13}l_{12}\varepsilon _{32}+\\&l_{11}l_{13}\varepsilon _{13}+l_{12}l_{13}\varepsilon _{23}+l_{13}l_{13}\varepsilon _{33}\\=&l_{11}(l_{11}\varepsilon _{11}+l_{12}\varepsilon _{12}+l_{13}\varepsilon _{13})+l_{12}(l_{11}\varepsilon _{21}+l_{12}\varepsilon _{22}+l_{13}\varepsilon _{23})+\\&l_{13}(l_{11}\varepsilon _{31}+l_{12}\varepsilon _{32}+l_{13}\varepsilon _{33})\\=&({\frac {\sqrt {3}}{2}})\left[({\frac {\sqrt {3}}{2}})(0.01)+({\frac {1}{2}})\varepsilon _{12}\right]+({\frac {1}{2}})\left[({\frac {\sqrt {3}}{2}})\varepsilon _{12}+({\frac {1}{2}})(0.02)\right]\\=&(3/4)(0.01)+({\sqrt {3}}/2)\varepsilon _{12}+(1/4)(0.02)\\=&(5/4)(0.01)+({\sqrt {3}}/2)\varepsilon _{12}\end{aligned}}}$

Therefore,

${\displaystyle (5/4)(0.01)+({\sqrt {3}}/2)\varepsilon _{12}=\varepsilon _{30^{o}}=0}$

Hence,

${\displaystyle \varepsilon _{12}=-(2.5)(0.01)/{\sqrt {3}}}$

Next, for an angle of rotation of 60${\displaystyle ^{o}}$, the matrix ${\displaystyle \left[L\right]}$ is

${\displaystyle {\begin{aligned}\left[L\right]&={\begin{bmatrix}\cos(60^{o})&\sin(60^{o})&\cos(90^{o})\\-\sin(60^{o})&\cos(60^{o})&\cos(90^{o})\\\cos(90^{o})&\cos(90^{o})&\cos(0^{o})\end{bmatrix}}\\&={\begin{bmatrix}1/2&{\sqrt {3}}/2&0\\-{\sqrt {3}}/2&1/2&0\\0&0&1\end{bmatrix}}\end{aligned}}}$

Therefore, ${\displaystyle \varepsilon _{60^{o}}=\varepsilon _{11}^{'}}$, is given by

${\displaystyle {\begin{aligned}\varepsilon _{11}^{'}=&l_{1p}l_{1q}\varepsilon _{pq}\\=&l_{11}l_{11}\varepsilon _{11}+l_{12}l_{11}\varepsilon _{21}+l_{13}l_{11}\varepsilon _{31}+l_{11}l_{12}\varepsilon _{12}+l_{12}l_{12}\varepsilon _{22}+l_{13}l_{12}\varepsilon _{32}+\\&l_{11}l_{13}\varepsilon _{13}+l_{12}l_{13}\varepsilon _{23}+l_{13}l_{13}\varepsilon _{33}\\=&l_{11}(l_{11}\varepsilon _{11}+l_{12}\varepsilon _{12}+l_{13}\varepsilon _{13})+l_{12}(l_{11}\varepsilon _{21}+l_{12}\varepsilon _{22}+l_{13}\varepsilon _{23})+\\&l_{13}(l_{11}\varepsilon _{31}+l_{12}\varepsilon _{32}+l_{13}\varepsilon _{33})\\=&({\frac {1}{2}})\left[({\frac {1}{2}})(0.01)+({\frac {\sqrt {3}}{2}})\varepsilon _{12}\right]+({\frac {\sqrt {3}}{2}})\left[({\frac {1}{2}})\varepsilon _{12}+({\frac {\sqrt {3}}{2}})(0.02)\right]\\=&(1/4)(0.01)+({\sqrt {3}}/2)\varepsilon _{12}+(3/4)(0.02)\\=&(7/4)(0.01)+({\sqrt {3}}/2)(-(2.5)(0.01)/{\sqrt {3}})\\=&(7/4)(0.01)-(5/4)(0.01)=(1/2)(0.01)=0.005\\\end{aligned}}}$

Therefore,

${\displaystyle {\varepsilon _{60^{o}}=0.005}}$

Part (b)

${\displaystyle {\text{The result is valid for all materials.}}}$