Thermal Shock and Residual Strength Testing of SiC/SiC Composite Braided Tubes

Abstract

Background

Ceramic matrix composites are promising materials for high temperature application in aerospace and nuclear engineering. In these applications, thermal shock is an important potential cause for failure.

Objective

In order to study thermal shock resistance of SiC/SiC composite braided tubes, a novel method has been developed to apply thermal shock cycles to tube sections and then measure the residual tensile strength.

Methods

SiC/SiC composite braided tubes have been thermally shocked by many cycles in a short time using a novel test platform based on quartz lamp irradiation heating. The circumferential tensile strength was measured using C-ring specimens after thermal shock testing of short tube sections. Numerical simulations of the stress from the thermal shock test were conducted using the finite element method.

Results

The circumferential tensile strength decreased with increasing number of thermal shock cycles in air. An embrittled region with limited fiber pullout due to oxidation extended from the surface.

Conclusions

The test platform can simulate service environments with fast temperature cycling for small test specimens in air.

Appendix

According to curved beam theory [37], the circumferential stress at position y is

$$\sigma_{\theta } = - \frac{F}{A} + \frac{M}{AR} - \frac{M}{{J_{z} }}\frac{y}{{1 - \frac{y}{R}}}$$

(3)

where R is radius (at the mid-thickness of the tube wall), F is the force, y is the coordinate in the Y direction (y=0 at the mid-thickness of the tube wall), as illustrated in Fig. 12, A is the cross-sectional area, and M is the bending moment.

Fig. 12

The defined geometry parameters of C-shaped specimen and cross-section A-A

The expression of J_Z is as follows:

$$J_{z} = \int\limits_{A} {\frac{{y^{2} {\text{d}}A}}{{(1 - \frac{y}{R})}}}$$

(4)

For the C-shaped specimen, failure tends to propagate from the symmetrical position of the specimen (i.e., A-A plane). On the A-A plane, the cross-sectional area and the bending moment are

$$A = bh$$

(5)

$$M = FR$$

(6)

For rectangular sections,

$$\begin{gathered} J_{z} = \int\limits_{A} {\frac{{y^{2} dA}}{{1 - \frac{y}{R}}} = \int\limits_{{ - \frac{h}{2}}}^{\frac{h}{2}} {(y^{2} + \frac{{y^{4} }}{{R^{2} }} + \frac{{y^{6} }}{{R^{4} }} + ...)} } b{\text{d}}y \\ = \frac{{bh^{3} }}{12}(1 + \frac{{3h^{2} }}{{20R^{2} }} + \frac{{3h^{4} }}{{112R^{4} }} + ...) \\ \\ \end{gathered}$$

(7)

It can be found that

$$J_{Z} = \left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {1.177I_{z} ,} & {R = h} \\ \end{array} } \\ {\begin{array}{*{20}c} {1.039I_{z} ,} & {R = 2h} \\ \end{array} } \\ {\begin{array}{*{20}c} {1.017I_{z} ,} & {R = 3h} \\ \end{array} } \\ {\begin{array}{*{20}c} {1.009I_{z} ,} & {R = 4h} \\ \end{array} } \\ \end{array} } \right.$$

(8)

Therefore, in order to simplify the calculation, it is assumed that J_Z≈I_Z, R=4.5 mm and h=1 mm. Then, circumferential stress at the peak load can be calculated as

$$\sigma_{\theta } = - \frac{12FR}{{bh^{3} }}\frac{y}{{1 - \frac{y}{R}}}$$

(9)

As can be seen from equation (9), when y=h/2, σ_θ has the maximum value σ_θmax.

$$\sigma_{\theta \max } = \frac{6FR}{{bh^{2} (1 + \frac{h}{2R})}}$$

(10)