Chapter 5
Q. 5.4
Repeat Example 5.3 for the first ply failure and use Tsai–Wu failure theory now with an additional thermal load: a temperature change of –75°C.
Step-by-Step
Verified Solution
The laminate is symmetric and the load applied is a normal load and a temperature change. Thus, only the extensional stiffness matrix is needed.
From Example 5.3,
\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} .
Corresponding to a temperature change of –75°C, the mechanical stresses can be found as follows. The fictitious thermal forces given by Equation (4.64) are
\left[\begin{array}{c} N_{x}^{T} \\ N_{y}^{T} \\ N_{x y}^{T} \end{array}\right]=(-75)\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} 0.200 \times 10^{-7} \\ 0.225 \times 10^{-4} \\ 0 \end{array}\right][-0.0025-(-0.0075)] \\ \space \\ +(-75)\left[\begin{array}{ccc} 10.35 & 2.897 & 0 \\ 2.897 & 181.8 & 0 \\ 0 & 0 & 7.17 \end{array}\right]\left(10^{9}\right)\left[\begin{array}{c} 0.225 \times 10^{-4} \\ 0.200 \times 10^{-7} \\ 0 \end{array}\right][0.0025-(-0.0025)] \\ \space \\ +(-75)\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} 0.200 \times 10^{-7} \\ 0.225 \times 10^{-4} \\ 0 \end{array}\right][0.0075-(-0.0025)] \\ \space \\ =\left[\begin{array}{c} -1.389 \times 10^{5} \\ -2.004 \times 10^{5} \\ 0 \end{array}\right] P a-m .
Because the laminate is symmetric, the fictitious thermal moments are zero.
This also then gives only midplane strains in the laminate without any plate curvatures. The midplane strain due to the thermal load is given by
\left[\begin{array}{c} \varepsilon_{x}^{0} \\ \varepsilon_{y}^{0} \\ \gamma_{x y}^{0} \end{array}\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]\left[\begin{array}{c} -1.389 \times 10^{5} \\ -2.004 \times 10^{5} \\ 0 \end{array}\right] \\ \space \\ =\left[\begin{array}{c} -0.6977 \times 10^{-4} \\ -0.1950 \times 10^{-3} \\ 0 \end{array}\right].
The laminate is symmetric and no bending or torsional moments are applied; therefore, the global strains in the laminate are the same as the midplane strains. The free expansional thermal strains in the top 0° ply are
\left[\begin{array}{l} \alpha_{x} \\ \alpha_{y} \\ \alpha_{x y} \end{array}\right]_{0^{\circ}}\Delta T\\ \space \\ =\left[\begin{array}{c} 0.200 \times 10^{-7} \\ 0.225 \times 10^{-4} \\ 0 \end{array}\right](-75) \\ \space \\ =\left[\begin{array}{c} -0.1500 \times 10^{-5} \\ -0.16875 \times 10^{-2} \\ 0 \end{array}\right].
From Equation (4.70), the global mechanical strain at the top surface of the top 0° ply is
\left[\begin{array}{c} -0.6977 \times 10^{-4} \\ -0.1950 \times 10^{-3} \\ 0 \end{array}\right]-\left[\begin{array}{c} -0.1500 \times 10^{-5} \\ -0.16875 \times 10^{-2} \\ 0 \end{array}\right] \\ \space \\ =\left[\begin{array}{c} -0.6827 \times 10^{-4} \\ 0.14925 \times 10^{-2} \\ 0 \end{array}\right]
From Equation (2.103), the global mechanical stresses at the top of the top 0° ply are
\left[\begin{array}{c} \sigma_{x} \\ \sigma_{y} \\ \tau_{x y} \end{array}\right]=\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} -0.6827 \times 10^{-4} \\ 0.14925 \times 10^{-2} \\ 0 \end{array}\right] \\ \space \\ =\left[\begin{array}{c} -8.088 \times 10^{6} \\ 1.524 \times 10^{7} \\ 0 \end{array}\right] P a.
Now, if the mechanical loads were given, the resulting mechanical stresses could then be added to the previous stresses due to the temperature difference.
Then, the failure criteria could be used to find out whether the ply has failed. However, we are asked to find out the mechanical load that could be applied in the presence of the temperature difference. This can be done as follows.
The stress at the top of the 0° ply, per Example 5.3 for a unit load N_x = 1 N/m, is
\left[\begin{array}{c} \sigma_{x} \\ \sigma_{y} \\ \tau_{x y} \end{array}\right]=\left[\begin{array}{c} 9.726 \times 10^{1} \\ 1.313 \times 10^{0} \\ 0.0 \end{array}\right] P a.
If the unknown load is N_x, then the overall stress at the top surface of the top 0° ply is
\left[\begin{array}{c} \sigma_{x} \\ \sigma_{y} \\ \tau_{x y} \end{array}\right]=\left[\begin{array}{c} -8.088 \times 10^{6}+9.726 \times 10^{1} N_{x} \\ 1.524 \times 10^{7}+1.313 \times 10^{0} N_{x} \\ 0 \end{array}\right] P a.
Now, the failure theories can be applied to find the value of N_x. Using transformation equation (2.94), the local stresses at the top surface of the top 0° ply are
\left[\begin{array}{c} \sigma_{1} \\ \sigma_{2} \\ \tau_{12} \end{array}\right]=\left[\begin{array}{c} -8.088 \times 10^{6}+9.726 \times 10^{1} N_{x} \\ 1.524 \times 10^{7}+1.313 \times 10^{0} N_{x} \\ 0 \end{array}\right] P a.
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 criterion (Equation 2.146) is
(0) [–8.088 × 10^6 + 9.726 × 10^1 N_x] + (2.093 × 10^{–8})[1.524 × 10^7 + 1.313 × 10^0 N_x]+(0)(0) + 4.4444 × 10^{–19} [–8.088 × 10^6 + 9.726 × 10^1 N_x]^2+ 1.0162 × 10^{–16} [1.524 × 10^7 + 1.313 × 10^0 N_x]^2 + 2.1626 × 10^{–16} [0]^2 + 2[–3.360 × 10^{–18}] [–8.088 × 10^6 + 9.726 × 10^1 N_x] [1.524 × 10^7 + 1.313× 10^0 N_x] < 1.
TABLE 5.7
Strength Ratios of Example 5.4
| Ply no. | Position | Tsai–Wu |
| 1 (0°) | Top | 1.100 × 10^7 |
| Middle | 1.100 × 10^7 | |
| Bottom | 1.100 × 10^7 | |
| 2 (90°) | Top | 4.279 × 10^6 |
| Middle | 4.279 × 10^6 | |
| Bottom | 4.279 × 10^6 | |
| 3 (0°) | Top | 1.100 × 10^7 |
| Middle | 1.100 × 10^7 | |
| Bottom | 1.100 × 10^7 |
As can be seen, this results in a quadratic polynomial in the left-hand side of the strength criteria — that is,
3.521 \times 10^{-15} N_{x}^{2}+2.096 \times 10^{-8} N_{x}-0.6566=0.
This gives two roots for which the inequality is satisfied for N_x < 1.100 × 10^7 and N_x > –1.695 × 10^7.
Because the load N_x is tensile, N_x = 1.100 × 10^7 is the valid solution.
Similarly, the values of strength ratios for all the plies in the laminate are found and summarized in Table 5.7.
Using the lowest value of strength ratio of 4.279 × 10^6 gives N_x = 4.279 × 10^6 N/m as the load at which the first ply failure would take place. Compare this with the value of N_x = 7.277 × 10^6 in Example 5.3, in which no temperature change was applied.
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(2.146): \left(G_{1}+G_{3}\right)\left(\sigma_{2}^{T}\right)_{u l t}^{2}=1