A symmetric single-ply laminate comprises four isotropic plies of which the two outer plies are each 0.15 mm thick and are of aluminium alloy material while the two inner plies are each 0.15 mm thick but are of mild steel material. The ply angle for the outer plies is +45° while that for the inner

Question

A symmetric single-ply laminate comprises four isotropic plies of which the two outer plies are each 0.15 mm thick and are of aluminium alloy material while the two inner plies are each 0.15 mm thick but are of mild steel material. The ply angle for the outer plies is +45° while that for the inner plies is -45°. If, for the aluminium alloy [latex]E_{ al }=70,000 N / mm ^2 ext { and } ν _{ al }=0.3[/latex] while for mild steel [latex]E_{\mathrm{st}}\!=\!200,000\,\mathrm{N/m}m^{2}\,\mathrm{and}\ \mathrm{ν_{st}}\!=\!0.3,[/latex] calculate the equivalent elastic constants for the laminate.

Accepted Answer

For the aluminium alloy plies [latex]E_{1}\!=\!E_{\mathrm{t}}\!=\!E=70000\;\mathrm{N/mm}^{2},\; u_{\mathrm{lt}}\!=\! u_{\mathrm{tl}}\!=\!0.3,[/latex] nd for an isotropic material, from Eq. (1.50),

[latex]G_{\mathrm{{lt}}}={\frac{70000}{2(1+0.3)}}=26923\mathrm{{N/mm^{2}}}[/latex]

Then, from Eq. (26.17),

[latex]\left\{\begin{array}{l}\sigma_x \\sigma_y \ au_{x y}\end{array} ight\}=\left[\begin{array}{ccc}\frac{E_l}{1-ν_{l t} ν_{t l}} & \frac{ν_{t l} E_l}{1-ν_{lt} 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} ight]\left\{\begin{array}{c}\varepsilon_l \\varepsilon_t \\gamma_{l t}\end{array} ight\}[/latex] (26.17)[latex]\begin{aligned}& k_{11}=70000 /\left(1-0.3^2 ight)=76923 N / mm ^2 \& k_{22}=76923 N / mm ^2 \& k_{33}=26923 N / mm ^2 \& k_{12}=0.3 imes 70000 /\left(1-0.3^2 ight)=23077 N / mm ^2 \& k_{13}=k_{23}=0\end{aligned}[/latex]

Since the ply angle is 45°, m=1/√2 =n. Then, from Eqs (26.31),

[latex]\begin{gathered}\left\{\begin{array}{c}\sigma_x \\sigma_y \ au_{x y}\end{array} ight\}=\left[\begin{array}{ccc}m^4k_{11}+m^2 n^2\left(2 k_{12} ight. & m^2n^2\left(k_{11}+k_{22}-4 k_{33} ight) & m^3n\left(k_{11}-k_{12}-2 k_{33} ight) \\left.+4k_{33} ight)+n^4 k_{22} &+\left(m^4+n^4 ight) k_{12} & +m n^3\left(k_{12}-k_{22}+2 k_{33} ight) \m^2 n^2\left(k_{11}+k_{22}-4k_{33} ight) & n^4 k_{11}+m^2 n^2\left(2 k_{12} ight. & m n^3\left(k_{11}-k_{12}-2 k_{33} ight) \+\left(m^4+n^4 ight) k_{12} &\left.+4 k_{33} ight)+m^4 k_{22} & +m^3n\left(k_{12}-k_{22}+2 k_{33} ight) \m^3 n\left(k_{11}-k_{12}-2 k_{33} ight) & m n^3\left(k_{11}-k_{12}-2 k_{33} ight) & m^2 n^2\left(k_{11}+k_{22}-2k_{12} ight. \+mn^3\left(k_{12}+k_{22}+2 k_{33} ight) & +m^3n\left(k_{12}-k_{22}+2 k_{33} ight) & \left.-2k_{33} ight)+\left(m^4+n^4 ight) k_{33}\end{array} ight]\left\{\begin{array}{c}\varepsilon_x \\varepsilon_y \\gamma_{x y}\end{array} ight\}\end{gathered}[/latex] (26.31)

[latex]k_{11}=(1 / \sqrt{2})^4 imes 76923+(1 /\sqrt{2})^2(1 / \sqrt{2})^2(2 imes 23077+4 imes 26923)+(1 / \sqrt{2})^4 imes 76923[/latex]

i.e.,

[latex]{\overline{{k}}}_{11}=76923N/{\mathrm{mm}^{2}}[/latex]

Similarly,

[latex]\begin{aligned}& \bar{k}_{22}=76923 N / mm ^2\& \bar{k}_{33}=26923 N / mm ^2 \&\bar{k}_{12}=23077 N / mm ^2 \&\bar{k}_{13}=0=\bar{k}_{23}\end{aligned}[/latex]

For the steel plies, [latex]E_1=E_{ t }=E=200000 N / mm ^2, u_{ lt }= u_{ tl }=0.3[/latex] and from Eq. (1.50),
G=200000/2(1 + 0.3)=76923 N/mm². Then, from Eq. (26.17),

[latex]\begin{aligned}& k_{11}=200000 /\left(1-0.3^2 ight)=219780 N / mm ^2 \& k_{22}=219780 N /mm ^2 \& k_{33}=76923 N / mm ^2 \& k_{12}=0.3 imes 200000 /\left(1-0.3^2 ight)=65934 N / mm ^2\& k_{13}=k_{23}=0\end{aligned}[/latex]

The ply angle is –45° so that m=1/√2 and n=-1/√2. Then, from Eqs (26.31),

[latex]\begin{aligned}& \bar{k}_{11}=219780 N / mm ^2\& \bar{k}_{22}=219780 N / mm ^2 \&\bar{k}_{33}=76923 N / mm ^2 \&\bar{k}_{12}=65934 N / mm ^2 \&\bar{k}_{13}=\bar{k}_{23}=0\end{aligned}[/latex]

From Eq. (26.47),

[latex]A_{ ij }=\sum\limits_{ p =1}^{ N } t_{ p }\left(\bar{k}_{ ij } ight)_{ p } \quad( i =1,2,3, \ldots, j =1,2,3, \ldots)[/latex] (26.47)

[latex]\begin{aligned}& A_{11}=2 imes 0.15(76923+219780)=89011 N / mm \ & A_{22}=2 imes 0.15(76923+219780)=89011 N / mm \& A_{33}=2 imes 0.15(26923+76923)=31154 N / mm \& A_{12}=2 imes 0.15(23077+65934)=26703 N / mm \& A_{13}=A_{23}=0\end{aligned}[/latex]

Then, from Eq. (26.54),

[latex]A A=\left(A_{11}A_{22}-A_{12}^{\ 2} ight)A_{33}[/latex] (26.54)

[latex]A A=89011 imes 89011 imes 31154-26703^2 imes 31154=2.25 imes 10^{14}[/latex]

Now, from Eqs (26.55),

[latex]\begin{aligned}& a_{11}=\left(A_{22} ight)/\left(A_{11} A_{22}-A_{12}^2 ight) \& a_{22}=\left(A_{11} ight)/\left(A_{11} A_{22}-A_{12}{ }^2 ight) \& a_{33}=1 / A_{33} \& a_{12}=-\left(A_{12} ight) /\left(A_{11} A_{22}-A_{12}{}^2 ight) \& a_{13}=0 \& a_{23}=0\end{aligned}[/latex] (26.55)[latex]\begin{aligned}& a_{11}=89011 imes 31154 / 2.25 imes 10^{14}=1.23 imes 10^{-5} \& a_{22}=1.23 imes 10^{-5} \& a_{33}=3.21 imes 10^{-5} \& a_{12}=-0.37 imes 10^{-5} \& a_{13}=a_{23}=0\end{aligned}[/latex]

From Eq. (26.56),

[latex]E_{x}={\frac{1}{t a_{11}}}[/latex] (26.56)

[latex]E_x=\frac{1}{4 imes 0.15 imes 1.23 imes 10^{-5}}=135501 N / mm ^2[/latex]

From Eq. (26.59),

[latex]E_{y}={\frac{1}{t a_{22}}}[/latex] (26.59)

[latex]E_y=\frac{1}{4 imes 0.15 imes 1.23 imes 10^{-5}}=135501 N / mm ^2[/latex]

From Eq. (26.62),

[latex]\frac{\bar{ au}_{x y}}{\gamma_{x y}}=\frac{1}{ta_{33}}=G_{x y}[/latex] (26.62)

[latex] G_{x y}=\frac{1}{4 imes 0.15 imes 3.21 imes 10^{-5}}=51921 N / mm ^2[/latex]

From Eq. (26.57),

[latex]ν_{x y}=\frac{-\varepsilon_y}{\varepsilon_x}=\frac{-a_{12}}{a_{11}}[/latex] (26.57)
[latex] u_{x y}=-\frac{\left(-0.37 imes 10^{-5} ight)}{1.23 imes 10^{-5}}=0.3[/latex]

From Eq. (26.60),

[latex]ν_{y x}=\frac{-a_{12}}{a_{22}}[/latex] (26.60)

[latex] u_{y x}=-\frac{\left(-0.37 imes 10^{-5} ight)}{1.23 imes 10^{-5}}=0.3[/latex]

Finally, from Eq. (26.58), [latex]m_{x}{=}0,[/latex] and from Eq. (26.61), [latex]m_{y}=0.[/latex]

[latex]\frac{\gamma_{x y}}{\varepsilon_x}=\frac{a_{13}}{a_{11}}=-m_x[/latex] (26.58)

[latex]m_{\mathrm{y}}={\frac{-a_{23}}{a_{22}}}[/latex] (26.61)

Question 26.10: A symmetric single-ply laminate comprises four isotropic pli......

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