Question 24.10: If the ply angle in the laminate of Example 24.9 is 45° calc...

The minor Poisson’s ratio has the same value, 0.021, as in Example 24.9. Also the reduced stiffness terms are the same as in Example 24.9, i.e., k_{13}=k_{23}=0, k_{11}=140888 N / mm ^{2}, k_{12}=2958 N / mm ^{2}, k_{22}=10060 N / mm ^{2} and k_{33}=5000 N / mm ^{2}. Now, however, \theta=45^{\circ} so that m=\cos 45^{\circ}=1 / \sqrt{2} and n=\sin 45^{\circ}=1 / \sqrt{2} Then, in Eq. (24.31)

\left\{\begin{array}{c}\sigma_{x} \\\sigma_{y} \\\tau_{x y}\end{array}\right\}=\left[\begin{array}{ccc}m^{4}k_{11}+m^{2}n^{2}\left(2k_{12}\right.&m^{2}n^{2}\left(k_{11}+k_{22}-4k_{33}\right)&m^{3}n\left(k_{11}-k_{12}-2k_{33}\right)\\\left.+4k_{33}\right)+n^{4}k_{22}&+\left(m^{4}+n^{4}\right)k_{12}&+mn^{3}\left(k_{12}-k_{22}+2k_{33}\right)\\m^{2}n^{2}\left(k_{11}+k_{22}-4k_{33}\right)&n^{4}k_{11}+m^{2}n^{2}\left(2k_{12}\right. & m n^{3}\left(k_{11}-k_{12}-2 k_{33}\right)\\+\left(m^{4}+n^{4}\right)k_{12}&\left.+4k_{33}\right)+m^{4} k_{22} & +m^{3} n\left(k_{12}-k_{22}+2 k_{33}\right)\\m^{3} n\left(k_{11}-k_{12}-2k_{33}\right)&mn^{3}\left(k_{11}-k_{12}-2k_{33}\right)&m^{2}n^{2}\left(k_{11}+k_{22}-2k_{12}\right.\\+mn^{3}\left(k_{12}+k_{22}+2k_{33}\right)& +m^{3} n\left(k_{12}-k_{22}+2k_{33}\right)&\left.-2k_{33}\right)+\left(m^{4}+n^{4}\right)k_{33}\end{array}\right]\left\{\begin{array}{c}\varepsilon_{x} \\\varepsilon_{y} \\\gamma_{x y}\end{array}\right\}  (24.31)

\bar{k}_{11}=m^{4} k_{11}+m^{2} n^{2}\left(2 k_{12}+4 k_{33}\right)+n^{4} k_{22}  (i)

Substituting the above values of k_{11} etc. in Eq. (i) gives

Similarly \bar{k}_{22}=44216 N / mm ^{2}

Then, from Eq. (24.45)

A_{11}=\sum\limits_{ p =1}^{ N }\left(z_{ p }-z_{ p -1}\right) \bar{k}_{11}  (24.45)

Similarly A_{22}=25296 N / mm

From Eq. (24.53)

A A=A_{11} A_{22} A_{33}+2 A_{12} A_{23} A_{13}-A_{22} A_{13}^{2}-A_{33} A_{12}^{2}-A_{11} A_{23}{ }^{2}  (24.53)

Then, from Eqs. (24.52)

a_{12}=\left(A_{13} A_{23}-A_{12} A_{33}\right) / A A  (24.52)

Similarly a_{12}=-0.57 \times 10^{-4}

From Eq. (24.56)

E_{x}=\frac{1}{t a_{11}}  (24.56)

From Eq. (24.59)

E_{y}=\frac{1}{t a_{22}}  (24.59)

From Eq. (24.62)

\frac{\bar{\tau}_{x y}}{\gamma_{x y}}=\frac{1}{t a_{33}}=G_{x y}  (24.62)

From Eq. (24.57)

ν_{x y}=\frac{-\varepsilon_{y}}{\varepsilon_{x}}=\frac{-a_{12}}{a_{11}}  (24.57)

From Eq. (24.60)

ν_{y x}=\frac{-a_{12}}{a_{22}}  (24.60)

From Eq. (24.58)

\frac{\gamma_{x y}}{\varepsilon_{x}}=\frac{a_{13}}{a_{11}}=-m_{x}  (24.58)

From Eq. (24.61)

m_{ y }=\frac{-a_{23}}{a_{22}}  (24.61)

Since the shear coupling coefficients have values it follows that, for this laminate, direct loads will produce shear strains and shear loads will produce direct strains. However, in many laminate configurations the ply angles are chosen such that the shear coupling effect is eliminated.