Many of the calculations needed to solve this problem can be completed using program CLT. Referring to the flow diagram shown in Figure 7.2:
Step 1: It is completed by the problem statement itself:
• Ply properties: listed in Table 3.1
• Laminate description: [0/30/60]_s
• Unit loads: As N_{xx} = k_1 = 1, and it is assumed that N_{yy}=k_2|N_{xx}|_\prime , N_{xy}=k_3|N_{xx}|_\prime , M_{xx}=k_4|N_{xx}|_\prime , M_{yy}=k_5|N_{xx}|_\prime , and M_{xy}=k_6|N_{xx}|_\prime , then k_1 = 1, k_2 = k_3 = k_4 = k_5 = k_6 = 0
• Temperature/moisture changes: ΔT = 20°C – 175°C = −155°C
ΔM = 0%
• Failure criterion: Maximum stress failure criterion
Step 2: Using ply properties for graphite–epoxy listed in Table 3.1 and program CLT, the [ABD] matrix for a [0/30/60]_s laminate is
[ ABD] = \left [ \begin{matrix} 76.44\times 10^6 & 13.79\times 10^6 & 17.41\times 10^6 & 0 & 0 & 0 \\13.79\times 10^6 & 36.23\times 10^6 & 17.41\times 10^6 & 0 & 0 & 0 \\ 17.41\times 10^6 & 17.41\times 10^6 &21.27\times 10^6 & 0 & 0 &0 \\ 0 & 0& 0 & 5.245 & 0.3461 & 0.4667 \\ 0 & 0 & 0 & 0.3461 & 0.6369 & 0.2588 \\ 0 & 0 & 0 & 0.4667 & 0.2588 & 0.6971 \end{matrix} \right ]
As an aside, the effective extensional stiffness of the laminate, based on the [ABD] matrix, is also returned by program CLT and equals \overline{E} _{xx}^{ex} = 82.9 GPa.
Step 3: Using ply properties for graphite–epoxy listed in Table 3.1, the specified change in temperature (ΔT = −155°C) and program CLT, the thermal stress and moment resultants are
\left\{ \begin{matrix} N_{xx}^T \\ N_{yy}^T \\ N_{xy}^T \\ M_{xx}^T \\ M_{yy}^T \\M_{xy}^T \end{matrix} \right\}=\left\{ \begin{matrix} -4803N/m \\ -18021 N/m\\ 11447 N/m \\ 0N- m/m \\ 0N- m/m \\0N- m/m \end{matrix} \right\}
As there is no change in moisture content, moisture stress and moment resultants are zero for this problem.
Step 4: Program CLT returns the ply stresses caused by the thermal stress and moment resultants for each ply, relative to the 1–2 coordinate system. These are listed in Table 7.1.
Step 5: On the basis of the specified unit loads (N_{xx} = k_1 = 1, k_2 = k_3 = k_4 = k_5 = k_6 = 0), and insuring that temperature and moisture changes are zero, program CLT returns the ply stresses relative to the 1–2 coordinate system listed in Table 7.1.
Step 6: The sum of steps 4 and 5 gives the stresses induced in each ply by the specified temperature change and any value of N_{xx}. For example, stresses induced in the 0° plies (i.e., plies 1 and 6) are given by
\sigma _{11}=2750N_{xx}-55.54\times 10^6(Pa)
\sigma _{22}=51.93N_{xx}+28.36\times 10^6(Pa)
\tau _{12}=-174.8N_{xx}+22.83\times 10^6(Pa)
The coefficients in these expressions are listed in Table 7.1. Analogous expressions for the stresses induced in the 30° and 60° plies can be constructed in the same way.
Step 7: The maximum stress failure criterion is to be applied in this problem. The criterion is applied to each ply in turn, using failure strengths listed in Table 3.1, thereby identifying the unit load factor necessary to cause failure of each ply.
TABLE 7.1
Numerical Results of the First-Ply Failure Analysis Described in Example Problem 7.1
|
Ply Stresses Caused by Temperature Change Only, 1–2 Coordinate System (MPa) |
Ply Stresses Caused by Unit Loads Only, 1–2 Coordinate System (Pa) |
|
| Ply Fiber Angle (°) |
σ_{11} |
σ_{22} |
\tau _{12} |
σ_{11} |
σ_{22} |
\tau _{12} |
Unit Load Scale Factor |
| 0 |
-55.54 |
28.36 |
22.83 |
2750 |
51.93 |
-174.8 |
416.7×10^3 |
| 30 |
29.59 |
24.79 |
0 |
1112 |
120.6 |
-264.6 |
209.0×10^3 |
| 60 |
-55.54 |
28.36 |
−22.83 |
−209.8 |
176·0 |
−89.85 |
123·0×10^3 |
From Table 3.1, failure strengths are
\sigma _{11}^{fT}=1500 MPa \sigma _{22}^{fT}=50 MPa \tau_{12}^{f}=\pm 90 MPa
The unit load factor necessary to cause failure of the 0° plies (i.e., plies 1 and 6) will be calculated to illustrate the process. Equating σ_{11} induced in the 0° plies by the temperature change and a unit load (determined in step 6), to the corresponding failure strength, we have
2750N_{xx}-55.54\times 10^6(Pa)=1500MPa\Rightarrow N_{xx}
=\frac{1500\times 10^6+55.54\times 10^6}{2750}=565.7KN/m
Thus, in order for a fiber failure to occur in the 0° plies, the tensile unit load must be increased by a load factor of 565.7 × 10^3. Repeating this process for stresses σ_{22} and τ_{12} induced in the 0 plies, we find
51.93N_{xx}+2836\times 10^6(Pa)=50MPa \Rightarrow N_{xx}=\frac{50\times 10^6-28.36\times 10^6}{51.93}
= 416.7 kN/m
174.8N_{xx}+22.83\times 10^6(Pa)=-90Pa \Rightarrow N_{xx}=\frac{-90\times 10^6-22.83\times 10^6}{-174.8}
=645.5KN/m
As failure of the 0° plies would occur for the lowest value of N_{xx} that causes any stress component to reach a critical level, we conclude that the critical load factor for the 0° plies equals 416.7 × 10^3.
Repeating this process for the 30° and 60° plies results in critical unit load factors of 209.0 × 10^3 and 123.0 × 10^3, respectively. Hence, according to the maximum stress criterion, the first-ply failure load for the laminate is N_{xx }= 123.0 kN/m, which corresponds to failure of the 60° plies (plies 3 and 4). Equivalently, the effective first-ply failure stress is
\overline{\sigma }_{xx}=\frac{N_{xx}}{t}=\frac{123.0KN/m}{6(0.125 mm)}=164MPa
* As discussed in Section 6.5, the stress-free temperature is likely to be lower than the cure temperature. For simplicity, it is assumed in this text that the final cure temperature defines the stress-free temperature.
