Show that the extensional stiffness matrix for a general N-ply quasi-isotropic laminate is given by [A] = [U1 U4 0 U4 U1 0 0 0 U_1-U_4/2]h. where U1 and U4 are the stiffness invariants given by Equation (2.132) and h is the thickness of the laminate. Also, find the in-plane engineering stiffness

Question

Show that the extensional stiffness matrix for a general N-ply quasi-isotropic laminate is given by

[latex] [A]=\left[\begin{array}{ccc} U_{1} & U_{4} & 0 \ U_{4} & U_{1} & 0 \ 0 & 0 & \frac{U_{1}-U_{4}}{2} \end{array} ight] h [/latex]. (5.8)

where [latex]U_1 and U_4[/latex] are the stiffness invariants given by Equation (2.132) and h is the thickness of the laminate. Also, find the in-plane engineering stiffness constants of the laminate.

(2.132):
[latex] \frac{1}{E_{y}}=\bar{S}_{22} \ \space \ =S_{11} s^{4}+\left(2 S_{12}+S_{66} ight) c^{2} s^{2}+S_{22} c^{4} \ \space \ =\frac{1}{E_{1}} s^{4}+\left(-\frac{2 v_{12}}{E_{1}}+\frac{1}{G_{12}} ight) c^{2} s^{2}+\frac{1}{E_{2}} c^{4}, [/latex]

Accepted Answer

From Equation (2.131a), for a general angle ply with angle θ,

[latex] \bar{Q}_{11}=U_{1}+U_{2} \operatorname{Cos} 2 heta+U_{3} \operatorname{Cos} 4 heta [/latex]. (5.9)

For the [latex]k^{th}[/latex] ply of the quasi-isotropic laminate with an angle [latex]θ_k[/latex],

[latex] \left(\bar{Q}_{11} ight)_{k}=U_{1}+U_{2} \operatorname{Cos} 2 heta_{k}+U_{3} \operatorname{Cos} 4 heta_{k}[/latex] (5.10)

where

[latex] heta_{1}=\frac{\pi}{N}, heta_{2}=\frac{2 \pi}{N}, \ldots, heta_{k}=\frac{k \pi}{N}, \ldots, heta_{N-1}=\frac{(N-1) \pi}{N}, heta_{N}=\pi . [/latex]

From Equation (4.28a),

[latex] A_{11}=\sum_{k=1}^{N} t_{k}\left(\bar{Q}_{11} ight)_{k} [/latex] (5.11)

where [latex]t_k[/latex] = thickness of [latex]k^{th}[/latex] lamina.
Because the thickness of the laminate is h and all laminae are of the same thickness,

[latex] t_{k}=\frac{h}{N}, \quad k=1,2, \ldots \ldots \ldots \ldots, N [/latex] (5.12)

and, substituting Equation (5.10) in Equation (5.11),

[latex] A_{11}=\frac{h}{N} \sum_{k=1}^{N}\left(U_{1}+U_{2} \operatorname{Cos} 2 heta_{k}+U_{3} \operatorname{Cos} 4 heta_{k} ight) \ \space \ =h U_{1}+U_{2} \frac{h}{N} \sum_{k=1}^{N} \operatorname{Cos} 2 heta_{k}+U_{3} \frac{h}{N} \sum_{k=1}^{N} \operatorname{Cos} 4 heta_{k} .[/latex] (5.13)

Using the following identity,[latex]^1[/latex]

[latex] \sum_{k=1}^{N} \operatorname{Cos} k x \equiv \frac{\operatorname{Sin}\left(N+\frac{1}{2} ight) x}{2 \operatorname{Sin}\left(\frac{x}{2} ight)}-\frac{1}{2}.[/latex] (5.14)

Then,

[latex] \sum_{k=1}^{N} \operatorname{Cos} 2 heta_{k}=0 ext { for } N \geq 1 [/latex] (5.15a)

[latex] \sum_{k=1}^{N} \operatorname{Cos} 4 heta_{k}=0 ext { for } N \geq 3 . [/latex] (5.15b)

Thus,

[latex] A_{11}=U_{1} h [/latex]. (5.16a)

Similarly, it can be shown that

[latex] A_{12}=U_{4} h [/latex], (5.16b)

[latex] A_{22}=U_{1} h [/latex], (5.16c)

[latex] A_{66}=\left(\frac{U_{1}-U_{4}}{2} ight) h. [/latex] (5.16d)

Therefore,

[latex] [A]=\left[\begin{array}{ccc} U_{1} & U_{4} & 0 \ U_{4} & U_{1} & 0 \ 0 & 0 & \frac{U_{1}-U_{4}}{2} \end{array} ight] h [/latex]. (5.17)

Because Equation (5.15b) is valid only for N ≥ 3, this proves that one needs at least three plies to make a quasi-isotropic laminate.
For a symmetric quasi-isotropic laminate, the extensional compliance matrix is given by

[latex] \left[A^{*} ight]=\frac{1}{h}\left[\begin{array}{ccc} \frac{U_{1}}{U_{1}^{2}-U_{4}^{2}} & -\frac{U_{4}}{U_{1}^{2}-U_{4}^{2}} & 0 \ -\frac{U_{4}}{U_{1}^{2}-U_{4}^{2}} & \frac{U_{1}}{U_{1}^{2}-U_{4}^{2}} & 0 \ 0 & 0 & \frac{2}{U_{1}-U_{4}} \end{array} ight] [/latex]. (5.18)

From the definitions of engineering constants given in Equations (4.35), (4.37), (4.39), (4.42), and (4.45), and using Equation (5.18), the elastic moduli of the laminate are independent of the angle of the lamina and are given by

[latex] E_{x}=E_{y}=E_{i s o}=\frac{1}{A_{11}^{*} h}=\frac{U_{1}^{2}-U_{4}^{2}}{U_{1}} [/latex], (5.19a)

[latex] G_{x y}=G_{i s o}=\frac{1}{A_{66}^{*} h}=\frac{U_{1}-U_{4}}{2} [/latex], (5.19b)

[latex] u_{x y}= u_{y x}= u_{i s o}=-\frac{A_{12}^{*}}{A_{22}^{*}}=\frac{U_{4}}{U_{1}} [/latex] (5.19c)

_________________________________

(2.131):
[latex] u_{x y}=-E_{x} \bar{S}_{12} \ \space \ =-E_{x}\left[S_{12}\left(s^{4}+c^{4} ight)+\left(S_{11}+S_{22}-S_{66} ight) s^{2} c^{2} ight] \ \space \ =E_{x}\left[\frac{v_{12}}{E_{1}}\left(s^{4}+c^{4} ight)-\left(\frac{1}{E_{1}}+\frac{1}{E_{2}}-\frac{1}{G_{12}} ight) s^{2} c^{2} ight][/latex]

(4.28a): [latex] A_{i j}=\sum_{k=1}^{n}\left[\left(\bar{Q}_{i j} ight) ight]_{k}\left(h_{k}-h_{k-1} ight), \quad i=1,2,6 ; \quad j=1,2,6, [/latex]

(4.35): [latex] E_{x} \equiv \frac{\sigma_{x}}{\varepsilon_{x}^{0}}=\frac{N_{x} / h}{A_{11}^{*} N_{x}}=\frac{1}{h A_{11}^{*}} [/latex]

(4.37): [latex]E_{y} \equiv \frac{\sigma_{y}}{\varepsilon_{y}^{0}}=\frac{N_{y} / h}{A_{22}^{*} N_{y}}=\frac{1}{h A_{22}^{*}} [/latex]

(4.39): [latex] G_{x y} \equiv \frac{ au_{x y}}{\gamma_{x y}^{0}}=\frac{N_{x y} / h}{A_{66}^{*} N_{x y}}=\frac{1}{h A_{66}^{*}} [/latex]

(4.42): [latex] u_{x y} \equiv-\frac{\varepsilon_{y}^{0}}{\varepsilon_{x}^{0}}=-\frac{A_{12}^{*} N_{x}}{A_{11}^{*} N_{x}}=-\frac{A_{12}^{*}}{A_{11}^{*}} [/latex]

(4.45): [latex] u_{y x} \equiv-\frac{\varepsilon_{x}^{0}}{\varepsilon_{y}^{0}} =-\frac{A_{12}^{*}}{A_{22}^{*}} \frac{N_{y}}{N_{y}} =-\frac{A_{12}^{*}}{A_{22}^{*}} [/latex]

Question 5.2: Show that the extensional stiffness matrix for a general N-p...

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