Question

b) Calculate the elements of the stiffhess and compliance matrices of a graphite epoxy unidirectional ply. The engineering constants are given as E,-148 x 10 NE9.65x10 N/m2, G 4.55 x 10N/m2, 151 12-0.3 and v,0.6

Answer 1

For transversely isotropic materials,

The stiffness matrix in principal direction is given by,

...............{i}
The compliance matrix in the principal direction is given by,

0 12V21 1-1 1-a 0 0 0 G12

..............................{ii}

001-G 攻0

The relation between $\nu_{12}$ and   is given by

/21

.........................................{iii}

12E1 21 E2

From {iii} can be calculated as,

/21

$\nu_{21} = \frac{\nu_{12}E_{2}}{E_{1}}$

ie.$\nu_{21} = \frac{0.3\times9.65 \times 10^{9}}{148\times10^{9}}$

20.01956

Now substituting values in {i} gives

148 0.3x9.65 1-03x0.01956 1-0.3x001956 0.01956x148 0 | x109у㎡ 4.55 1-0.3x0.01956 1-0.3x 0.01965 0

$[Q] = \begin{bmatrix} 148.874&2.912& 0 \\2.912&9.707&0 \\ 0& 0 & 4.55 \end{bmatrix} \times 10^{9}N/m^{2}$

$[S] = \begin{bmatrix} \frac{1}{148}& \frac{-0.3}{148} & 0 \\ \frac{-0.01956}{9.65} &\frac{1}{9.65} &0 \\ 0& 0 & \frac{1}{4.55 }\end{bmatrix}\times 10^{-9} m^{2}/N$

$[S] = \begin{bmatrix}0.006757& -0.002027& 0 \\ -0.002027&0.1036 &0 \\ 0& 0 & 0.21978\end{bmatrix}\times 10^{-9} m^{2}/N$