Question 24.11: The symmetric four-ply laminate whose cross-section is shown...

Consider plies 1 and 4. The minor Poisson’s ratio is found using Eq. (24.14), i.e.,

 

\frac{ν_{t l}}{E_{t}}=\frac{ν_{l t}}{E_{l}}  (24.14)

 

ν_{t l}=\frac{15000}{200000} \times 0.3=0.023

 

As in Example 24.9 the reduced stiffness terms k_{13} and k_{23} are zero. Then, from Eq. (24.17)

 

\left\{\begin{array}{c}\sigma_{x} \\\sigma_{y} \\\tau_{x y}\end{array}\right\}=\left[\begin{array}{ccc}\frac{E_{l}}{1-ν_{l t} ν_{t l}} & \frac{ν_{t l} E_{l}}{1-ν_{l t} ν_{t l}} &0\\\frac{ν_{l t} E_{t}}{1-ν_{l t} ν_{t l}} & \frac{E_{t}}{1-ν_{l t} ν_{t l}}&0\\0 & 0 & G_{l t}\end{array}\right]\left\{\begin{array}{c}\varepsilon_{l}\\\varepsilon_{t}\\\gamma_{l t}\end{array}\right\}  (24.17)

 

k_{11}=200000 /(1-0.3 \times 0.023)=201410 N / mm ^{2}

 

k_{12}=4632 N / mm ^{2}

 

Similarly k_{22}=15106 N / mm ^{2}

 

k_{33}=10000 N / mm ^{2}

 

For plies 1 and 4, \theta=30^{\circ} so that m=\cos 30^{\circ}=0.866 and n=\sin 30^{\circ}=0.5 \text {. } Then from Eq. (24.31) (see Eq. (i) of Example 24.10)

 

\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)

 

\bar{k}_{11}=123583 N / mm ^{2}

 

\bar{k}_{12}=35908 N / mm ^{2}

 

\bar{k}_{13}=58436 N / mm ^{2}

 

\bar{k}_{22}=30431 N / mm ^{2}

 

\bar{k}_{23}=22232 N / mm ^{2}

 

\bar{k}_{33}=41266 N / mm ^{2}

 

The above procedure is repeated for plies 2 and 3 \left(\theta=45^{\circ}\right) and gives the following results.

 

ν_{t l}=0.021

 

k_{11}=140888 N / mm ^{2}

 

k_{12}=2958 N / mm ^{2}

 

k_{22}=10060 N / mm ^{2}

 

k_{33}=5000 N / mm ^{2}

 

\bar{k}_{11}=44216 N / mm ^{2}

 

\bar{k}_{12}=34216 N / mm ^{2}

 

\bar{k}_{13}=32707 N / mm ^{2}

 

Then  \bar{k}_{22}=44216 N / mm ^{2}

 

\bar{k}_{23}=32707 N / mm ^{2}

 

\bar{k}_{33}=36258 N / mm ^{2}

 

The above results are now combined to produce the A terms in the [A] matrix for the complete laminate. From Eq. (24.45)

 

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

 

A_{11}=2 \times 0.13(123583+44216)=43628 N / mm

 

A_{12}=2 \times 0.13(35908+34216)=18232 N / mm

 

A_{13}=2 \times 0.13(58436+32707)=23697 N / mm

 

A_{22}=2 \times 0.13(30431+44216)=19408 N / mm

 

A_{23}=2 \times 0.13(22232+32707)=14284 N / mm

 

A_{33}=2 \times 0.13(41266+36258)=20156 N / mm

 

Then, 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)

 

A A=43628 \times 19408 \times 20156+2 \times 18232 \times 14284 \times 23697-19408 \times 23697^{2}-20156 \times 18232^{2} -43628 \times 14284^{2}

 

\text { i.e., } A A=2.91 \times 10^{12}

 

Substituting the above in Eqs. (24.52) gives

 

a_{11}=\left(A_{22} A_{33}-A_{23}^{2}\right) / A A

 

a_{22}=\left(A_{11} A_{33}-A_{13}^{2}\right) / A A

 

a_{33}=\left(A_{11} A_{22}-A_{12}^{2}\right) / A A

 

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

 

a_{13}=\left(A_{12} A_{23}-A_{22} A_{13}\right) / A A

 

a_{23}=\left(A_{12} A_{13}-A_{11} A_{23}\right) / A A

 

a_{11}=6.43 \times 10^{-5}

 

a_{22}=10.92 \times 10^{-5}

 

a_{33}=17.67 \times 10^{-5}

 

a_{12}=-0.10 \times 10^{-5}

 

a_{13}=-6.85 \times 10^{-5}

 

a_{23}=-6.57 \times 10^{-5}

 

Now, from Eq. (24.56)

 

E_{x}=\frac{1}{t a_{11}}  (24.56)

 

E_{x}=\frac{1}{4 \times 0.13 \times 6.43 \times 10^{-5}}=29908 N / mm ^{2}

 

From Eq. (24.59)

 

E_{y}=\frac{1}{t a_{22}}  (24.59)

 

E_{y}=\frac{1}{4 \times 0.13 \times 10.92 \times 10^{-5}}=17611 N / mm ^{2}

 

From Eq. (24.62)

 

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

 

G_{ xy }=\frac{1}{4 \times 0.13 \times 17.67 \times 10^{-5}}=10883 N / mm ^{2}

 

From Eq. (24.57)

 

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

 

ν_{x y}=\frac{-\left(-0.10 \times 10^{-5}\right)}{6.43 \times 10^{-5}}=0.016

 

From Eq. (24.60)

 

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

 

ν_{y x}=\frac{-\left(-0.10 \times 10^{-5}\right)}{10.92 \times 10^{-5}}=0.009

 

From Eq. (24.58)

 

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

 

m_{x}=\frac{-\left(-6.85 \times 10^{-5}\right)}{6.43 \times 10^{-5}}=1.07

 

From Eq. (24.61)

 

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

 

m_{y}=\frac{-\left(6.85 \times 10^{-5}\right)}{10.92 \times 10^{-5}}=0.64