Products

## Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

## Holooly Tables

All the data tables that you may search for.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 5.3

Find the ply-by-ply failure loads for a $[0/\overline{90}]_s$ graphite/epoxy laminate. Assume the thickness of each ply is 5 mm and use properties of unidirectional graphite/epoxy lamina from Table 2.1. The only load applied is a tensile normal load in the x-direction — that is, the direction parallel to the fibers in the 0° ply.

TABLE 2.1
Typical Mechanical Properties of a Unidirectional Lamina (SI System of Units)

 Property Symbol Units Glass/ epoxy Boron/ epoxy Graphite/ epoxy Fiber volume fraction $V_f$ 0.45 0.50 0.70 Longitudinal elastic modulus $E_1$ GPa 38.6 204 181 Transverse elastic modulus $E_2$ GPa 8.27 18.50 10.30 Major Poisson’s ratio $\nu_{12}$ 0.26 0.23 0.28 Shear modulus $G_{12}$ GPa 4.14 5.59 7.17 Ultimate longitudinal tensile strength $(\sigma_1^T)_{ult}$ MPa 1062 1260 1500 Ultimate longitudinal compressive strength $(\sigma_1^C)_{ult}$ MPa 610 2500 1500 Ultimate transverse tensile strength $(\sigma_2^T)_{ult}$ MPa 31 61 40 Ultimate transverse compressive strength $(\sigma_2^C)_{ult}$ MPa 118 202 246 Ultimate in-plane shear strength $(\tau_{12})_{ult}$ MPa 72 67 68 Longitudinal coefficient of thermal expansion $\alpha_1$ μm/m/°C 8.6 6.1 0.02 Transverse coefficient of thermal expansion $\alpha_2$ μm/m/°C 22.1 30.3 22.5 Longitudinal coefficient of moisture expansion $\beta_1$ m/m/kg/kg 0.00 0.00 0.00 Transverse coefficient of moisture expansion $\beta_2$ m/m/kg/kg 0.60 0.60 0.60

## Verified Solution

Because the laminate is symmetric and the load applied is a normal load, only the extensional stiffness matrix is required. From Example 4.4, the extensional compliance matrix is

$\left[A^{*}\right]=\left[\begin{array}{ccc} 5.353 \times 10^{-10} & -2.297 \times 10^{-11} & 0 \\ -2.297 \times 10^{-11} & 9.886 \times 10^{-10} & 0 \\ 0 & 0 & 9.298 \times 10^{-9} \end{array}\right] \frac{1}{P a-m},$

which, from Equation (5.1a), gives the midplane strains for symmetric laminates subjected to $N_x = 1 N/m$ as

$\left[\begin{array}{c} \varepsilon_{x}^{0} \\ \varepsilon_{y}^{0} \\ \gamma_{x y}^{0} \end{array}\right]=\left[\begin{array}{c} 5.353 \times 10^{-10} \\ -2.297 \times 10^{-11} \\ 0 \end{array}\right] .$

The midplane curvatures are zero because the laminate is symmetric and no bending and no twisting loads are applied.
The global strains in the top 0° ply at the top surface can be found as follows using Equation (4.16),

$\left[\begin{array}{l} \varepsilon_{x} \\ \varepsilon_{y} \\ \gamma_{x y} \end{array}\right]=\left[\begin{array}{c} 5.353 \times 10^{-10} \\ -2.297 \times 10^{-11} \\ 0 \end{array}\right]+(0.0075)\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\ \space \\ =\left[\begin{array}{c} 5.353 \times 10^{-10} \\ -2.297 \times 10^{-11} \\ 0 \end{array}\right]$.

Using Equation (2.103), one can find the global stresses at the top surface of the top 0° ply as

$\left[\begin{array}{c} \sigma_{x} \\ \sigma_{y} \\ \tau_{x y} \end{array}\right]_{0^{\circ} \text {,top }}=\left[\begin{array}{ccc} 181.8 & 2.897 & 0 \\ 2.897 & 10.35 & 0 \\ 0 & 0 & 7.17 \end{array}\right]\left(10^{9}\right)\left[\begin{array}{c} 5.353 \times 10^{-10} \\ -2.297 \times 10^{-11} \\ 0 \end{array}\right] \\ \space \\ =\left[\begin{array}{c} 9.726 \times 10^{1} \\ 1.313 \\ 0 \end{array}\right] P a.$

Using the transformation Equation (2.94), the local stresses at the top surface of the top 0° ply are

$\left[\begin{array}{l} \sigma_{1} \\ \sigma_{2} \\ \tau_{12} \end{array}\right]_{0^{\circ}, \text { top }}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 9.726 \times 10^{1} \\ 1.313 \times 10^{0} \\ 0 \end{array}\right] \\ \space \\ =\left[\begin{array}{c} 9.726 \times 10^{1} \\ 1.313 \\ 0 \end{array}\right] P a.$

All the local stresses and strains in the laminate are summarized in Table 5.1 and Table 5.2.

TABLE 5.1
Local Stresses (Pa) in Example 5.3

 Ply no. Position $σ_1$ $σ_2$ $τ_{12}$ 1 (0°) Top 9.726 × $10^1$ 1.313 × $10^0$ 0.0 Middle 9.726 × $10^1$ 1.313 × $10^0$ 0.0 Bottom 9.726 × $10^1$ 1.313 × $10^0$ 0.0 2 (90°) Top –2.626 × $10^0$ 5.472 × $10^0$ 0.0 Middle –2.626 × $10^0$ 5.472 × $10^0$ 0.0 Bottom –2.626 × $10^0$ 5.472 × $10^0$ 0.0 3 (0°) Top 9.726 × $10^1$ 1.313 × $10^0$ 0.0 Middle 9.726 × $10^1$ 1.313 × $10^0$ 0.0 Bottom 9.726 × $10^1$ 1.313 × $10^0$ 0.0

TABLE 5.2
Local Strains in Example 5.3

 Ply no. Position $ε_1$ $ε_2$ $τ_{12}$ 1 (0°) Top 5.353 × $10^{-10}$ –2.297 × $10^{-11}$ 0.0 Middle 5.353 × $10^{-10}$ –2.297 × $10^{-11}$ 0.0 Bottom 5.353 × $10^{-10}$ –2.297 × $10^{-11}$ 0.0 2 (90°) Top –2.297 × $10^{-11}$ 5.353 × $10^{-10}$ 0.0 Middle –2.297 × $10^{-11}$ 5.353 × $10^{-10}$ 0.0 Bottom –2.297 × $10^{-11}$ 5.353 × $10^{-10}$ 0.0 3 (0°) Top 5.353 × $10^{-10}$ –2.297 × $10^{-11}$ 0.0 Middle 5.353 × $10^{-10}$ –2.297 × $10^{-11}$ 0.0 Bottom 5.353 × $10^{-10}$ –2.297 × $10^{-11}$ 0.0

The Tsai–Wu failure theory applied to the top surface of the top 0° ply is applied as follows. The local stresses are

$σ_1 = 9.726 × 10^1 Pa \\ \space \\ σ_2 = 1.313 Pa \\ \space \\ τ_{12} = 0$

Using the parameters $H_1, H_2, H_6, H_{11}, H_{22}, H_{66}, and H_{12}$ from Example 2.19, the Tsai–Wu failure theory Equation (2.152) gives the strength ratio as

$(0) (9.726 × 10^1) SR + (2.093 × 10^{–8}) (1.313) SR+ (0 × 0) + (4.4444 × 10^{–19}) (9.726 × 10^1)^2(SR)^2 + (1.0162 × 10^{–16}) (1.313)^2(SR)^2 + (2.1626 × 10^{–16}) (0)^2 + 2(–3.360 × 10^{–18}) (9.726 × 10^1) (1.313)(SR)^2=1 \\ SR = 1.339 × 10^7.$

The maximum strain failure theory can also be applied to the top surface of the top 0° ply as follows. The local strains are

$\left[\begin{array}{l} \varepsilon_{1} \\ \varepsilon_{2} \\ \gamma_{12} \end{array}\right]=\left[\begin{array}{c} 5.353 \times 10^{-10} \\ -2.297 \times 10^{-11} \\ 0.000 \end{array}\right] .$

Then, according to maximum strain failure theory (Equation 2.143), the strength ratio is given by

$SR = \min \{[(1500 × 10^6)/(181 × 10^9)]/(5.353 × 10^{–10}),[(246 × 10^6)/(10.3 × 10^9)]/(2.297 × 10^{–11})\} = 1.548 × 10^7.$

The strength ratios for all the plies in the laminate are summarized in Table 5.3 using the maximum strain and Tsai–Wu failure theories. The symbols in the parentheses in the maximum strain failure theory column denote the mode of failure and are explained at the bottom of Table 2.3.

TABLE 5.3
Strength Ratios in Example 5.3

 Ply no. Position Maximum strain Tsai–Wu 1 (0°) Top 1.548 × $10^{7}$ (1T) 1.339 × $10^{7}$ Middle 1.548 × $10^{7}$ (1T) 1.339 × $10^{7}$ Bottom 1.548 × $10^{7}$ (1T) 1.339 × $10^{7}$ 2 (90°) Top 7.254 × $10^{6}$ (2T) 7.277 × $10^{6}$ Middle 7.254 × $10^{6}$ (2T) 7.277 × $10^{6}$ Bottom 7.254 × $10^{6}$ (2T) 7.277 × $10^{6}$ 3 (0°) Top 1.548 × $10^{7}$ (1T) 1.339 × $10^{7}$ Middle 1.548 × $10^{7}$ (1T) 1.339 × $10^{7}$ Bottom 1.548 × $10^{7}$ (1T) 1.339 × $10^{7}$

From Table 5.3 and using the Tsai–Wu theory, the minimum strength ratio is found for the 90° ply. This strength ratio gives the maximum value of the allowable normal load as

$N_{x}=7.277 \times 10^{6} \frac{ N }{ m }$

and the maximum value of the allowable normal stress as

$\frac{N_{x}}{h}=\frac{7.277 \times 10^{6}}{0.015}, \\ \space \\ =0.4851\times10^9 Pa$

where h = thickness of the laminate.
The normal strain in the x-direction at this load is

$\left(\varepsilon_{x}^{0}\right)_{\text {first ply failure }}\left(5.353 \times 10^{-10}\right)\left(7.277 \times 10^{6}\right) \\ \space \\ =3.895\times 10^{-3}$

Now, degrading the 90° ply completely involves assuming zero stiffnesses and strengths of the 90° lamina. Complete degradation of a ply does not allow further failure of that ply. For the undamaged plies, the $[0/\overline{90}]_s$ laminate has two reduced stiffness matrices as

$[Q]=\left[\begin{array}{ccc} 181.8 & 2.897 & 0 \\ 2.897 & 10.35 & 0 \\ 0 & 0 & 7.17 \end{array}\right] G P a$

and, for the damaged ply,

$[Q]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] G P a.$

Using Equation (4.28a), the extensional stiffness matrix

$A_{i j}=\sum_{k=1}^{3}\left[\bar{Q}_{i j}\right]_{k}\left(h_{k}-h_{k-1}\right) \\ \space \\ [A]=\left[\begin{array}{ccc} 181.8 & 2.897 & 0 \\ 2.897 & 10.35 & 0 \\ 0 & 0 & 7.17 \end{array}\right]\left(10^{9}\right)(0.005) \\ \space \\ +\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\left(10^{9}\right)(0.005) \\ \space \\ +\left[\begin{array}{ccc} 181.8 & 2.897 & 0 \\ 2.897 & 10.35 & 0 \\ 0 & 0 & 7.17 \end{array}\right]\left(10^{9}\right)(0.005) \\ \space \\ [A]=\left[\begin{array}{ccc} 181.8 & 2.897 & 0 \\ 2.897 & 10.35 & 0 \\ 0 & 0 & 7.17 \end{array}\right]\left(10^{7}\right) Pa – m.$

Inverting the new extensional stiffness matrix [A], the new extensional compliance matrix is

$\left[A^{*}\right]=\left[\begin{array}{ccc} 5.525 \times 10^{-10} & -1.547 \times 10^{-10} & 0 \\ -1.547 \times 10^{-10} & 9.709 \times 10^{-9} & 0 \\ 0 & 0 & 1.395 \times 10^{-8} \end{array}\right] \frac{1}{P a-m} ,$

which gives midplane strains subjected to $N_x = 1 N/m$ by Equation (5.1a) as

$\left[\begin{array}{c} \varepsilon_{x}^{0} \\ \varepsilon_{y}^{0} \\ \gamma_{x y} \end{array}\right]=\left[\begin{array}{ccc} 5.525 \times 10^{-10} & -1.547 \times 10^{-10} & 0 \\ -1.547 \times 10^{-10} & 9.709 \times 10^{-9} & 0 \\ 0 & 0 & 1.395 \times 10^{-8} \end{array}\right]\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]$

TABLE 5.4
Local Stresses after First Ply Failure in Example 5.3

 Ply no. Position $σ_1$ $σ_2$ $τ_{12}$ 1 (0°) Top 1.0000 × $10^{2}$ 0.0 0.0 Middle 1.0000 × $10^{2}$ 0.0 0.0 Bottom 1.0000 × $10^{2}$ 0.0 0.0 2 (90°) Top — — — Middle — — — Bottom — — — 3 (0°) Top 1.0000 × $10^{2}$ 0.0 0.0 Middle 1.0000 × $10^{2}$ 0.0 0.0 Bottom 1.0000 × $10^{2}$ 0.0 0.0

TABLE 5.5
Local Strains after First Ply Failure in Example 5.3

 Ply no. Position $ε_1$ $ε_2$ $γ_{12}$ 1 (0°) Top 5.25 × $10^{-10}$ –1.547 × $10^{-10}$ 0.0 Middle 5.525 × $10^{-10}$ –1.547 × $10^{-10}$ 0.0 Bottom 5.525 × $10^{-10}$ –1.547 × $10^{-10}$ 0.0 2 (90°) Top — — — Middle — — — Bottom — — — 3 (0°) Top 5.525 × $10^{-10}$ –1.547 × $10^{-10}$ 0.0 Middle 5.525 × $10^{-10}$ –1.547 × $10^{-10}$ 0.0 Bottom 5.525 × $10^{-10}$ –1.547 × $10^{-10}$ 0.0

$\left[\begin{array}{c} \varepsilon_{x}^{0} \\ \varepsilon_{y}^{0} \\ \gamma_{x y}^{0} \end{array}\right]=\left[\begin{array}{c} 5.525 \times 10^{-10} \\ -1.547 \times 10^{-10} \\ 0 \end{array}\right] .$

These strains are close to those obtained before the ply failure only because the 90° ply takes a small percentage of the load out of the normal load in the x-direction.
The local stresses in each layer are found using earlier techniques given in this example and are shown in Table 5.4. The strength ratios in each layer are also found using methods given in this example and are shown in Table 5.5.
From Table 5.6 and using Tsai–Wu failure theory, the minimum strength ratio is found in both the 0° plies. This strength ratio gives the maximum value of the normal load as

$N_{x}=1.5 \times 10^{7} \frac{N}{m}$

TABLE 5.6
Strength Ratios after First Ply Failure in Example 5.3

 Ply no. Position Maximum strain Tsai–Wu 1 (0°) Top 1.5000 × $10^{7}$ (1T) 1.5000 × $10^{7}$ Middle 1.5000 × $10^{7}$ (1T) 1.5000 × $10^{7}$ Bottom 1.5000 × $10^{7}$ (1T) 1.5000 × $10^{7}$ 2 (90°) Top — — Middle — — Bottom — — 3 (0°) Top 1.5000 × $10^{7}$ (1T) 1.5000 × $10^{7}$ Middle 1.5000 × $10^{7}$ (1T) 1.5000 × $10^{7}$ Bottom 1.5000 × $10^{7}$ (1T) 1.5000 × $10^{7}$

and the maximum value of the allowable normal stress as

$\frac{N_{x}}{h}=\frac{1.5 \times 10^{7}}{0.015} \\ \space \\ =1.0\times 10^9 Pa$

where h is the thickness of the laminate.
The normal strain in the x-direction at this load is

$\left(\varepsilon_{x}^{0}\right)_{\text {last ply failure }}=\left(5.525 \times 10^{-10}\right)\left(1.5 \times 10^{7}\right) \\ \space \\ =8.288\times 10^{-3}.$

The preceding load is also the last ply failure (LPF) because none of the layers is left undamaged. Plotting the stress vs. strain curve for the laminate until last ply failure shows that the curve will consist of two linear curves, each ending at each ply failure. The slope of the two lines will be the Young’s modulus in x direction for the undamaged laminate and for the FPF laminate — that is, using Equation (4.35),

$E_{x}=\frac{1}{(0.015)\left(5.353 \times 10^{-10}\right)} \\ \space \\ =124.5 GPa$

until first ply failure, and

$E_{x}=\frac{\left(N_{x} / h\right)_{\text {last ply failure }}-\left(N_{x} / h\right)_{\text {first play failure }}}{\left(\varepsilon_{x}^{o}\right)_{\text {last play failure }}-\left(\varepsilon_{x}^{o}\right)_{\text {first play failure }}} \\ \space \\ =\frac{0.1 \times 10^{10}-0.4851 \times 10^{9}}{8.288 \times 10^{-3}-3.895 \times 10^{-3}} \\ \space \\ = 117.2 GPa$

after first ply failure and until last ply failure (Figure 5.1).

____________________________________

(2.94):              $\left[\begin{array}{l} \sigma_{x} \\ \sigma_{y} \\ \tau_{x y} \end{array}\right]=[T]^{-1}\left[\begin{array}{l} \sigma_{1} \\ \sigma_{2} \\ \tau_{12} \end{array}\right]$

(2.103):          $\left[\begin{array}{l} \sigma_{x} \\ \sigma_{y} \\ \tau_{x y} \end{array}\right]=\left[\begin{array}{lll} \bar{Q}_{11} & \bar{Q}_{12} & \bar{Q}_{16} \\ \bar{Q}_{12} & \bar{Q}_{22} & \bar{Q}_{26} \\ \bar{Q}_{16} & \bar{Q}_{26} & \bar{Q}_{66} \end{array}\right]\left[\begin{array}{c} \varepsilon_{x} \\ \varepsilon_{y} \\ \gamma_{x y} \end{array}\right],$

(2.143):

$-\left(\varepsilon_{1}^{C}\right)_{u l t}<\varepsilon_{1}<\left(\varepsilon_{1}^{T}\right)_{u l t}, \text { or } \\ \space \\ -\left(\varepsilon_{2}^{C}\right)_{u l t}<\varepsilon_{2}<\left(\varepsilon_{2}^{T}\right)_{u l t}, \text { or } \\ \space \\ -\left(\gamma_{12}\right)_{u l t}<\gamma_{12}<\left(\gamma_{12}\right)_{u l t}$

(2.152):            $H_{1} \sigma_{1}+H_{2} \sigma_{2}+H_{6} \tau_{12}+H_{11} \sigma_{1}^{2}+H_{22} \sigma_{2}^{2}+H_{66} \tau_{12}^{2}+2 H_{12} \sigma_{1} \sigma_{2}<1$

(4.16):            $\left\{\begin{array}{c} \varepsilon_{x} \\ \varepsilon_{y} \\ \gamma_{x y} \end{array}\right\}=\left\{ \begin{array}{c} \varepsilon_{x}^{0} \\ \varepsilon_{y}^{0} \\ \gamma_{x y}^{0} \end{array}\right\}+z\left\{\begin{array}{c} \kappa_{x} \\ \kappa_{y} \\ \kappa_{x y} \end{array}\right\}$

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

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

(5.1a):                    $\left[\begin{array}{c} N_{x} \\ N_{y} \\ N_{x y} \end{array}\right]=\left[\begin{array}{lll} A_{11} & A_{12} & A_{16} \\ A_{12} & A_{22} & A_{26} \\ A_{16} & A_{26} & A_{66} \end{array}\right]\left[\begin{array}{c} \varepsilon_{x}^{0} \\ \varepsilon_{y}^{0} \\ \gamma_{x y}^{0} \end{array}\right]$

TABLE 2.3
Effect of Sign of Shear Stress as a Function of Angle of Lamina

 Angle, Degrees Positive $τ_{xy}$ MPa Negative $τ_{xy}$ MPa Shear strength MPa 0 68.00 (S) 68.00 (S) 68.00 15 78.52 (S) 78.52 (S) 78.52 30 136.0 (S) 46.19 (2T) 46.19 45 246.0 (2C) 40.00 (2T) 40.00 60 136.0 (S) 46.19 (2T) 46.19 75 78.52 (S) 78.52 (S) 78.52 90 68.00 (S) 68.00 (S) 68.00