應力強度因子(Stress Intensity Factor)K為一評估材料強度的重要指標，本文根據複合材料力學及線性破裂力學的原理，撰寫成程式計算出模型的邊界條件，並利用『ANSYS』工程分析軟體進行模型建立，模擬正交性複合材料存有一垂直纖維方向的張開型(Mode-I)裂縫，假定巨觀時的應力強度因子為1，來求相對微觀尺度下裂縫前端位於兩材質不同位置處的無因次應力強度因子K，探討不同的基材與纖維材料彈性模數比、體積分率比及主要波松比(Poisson’s Ratio)等狀態下，其應力強度因子是如何受到非均質材料之影響。
由文中結果可知，不論纖維體積分率為何，當裂縫端位於基材處
的K值皆比位於纖維處來的小，且隨著彈性模數比值增加，其差值也愈大。並清楚觀察到彈性模數比值的改變，對於裂縫端處於基材處各位置分析點K值影響較小，而對於主要承受外力的纖維處則是影響較大。

Stress Intensity Factor (SIF) is a vital target of material intensity in evaluating. This work is in terms of the principle of composite material mechanics and linear fracture mechanics, writing the program to calculate the boundary conditions of the model and simulating there is a Mode-I crack in vertical fiber direction of the orthotropic composite material, which is utilizing ANSYS engineering analysis software to build the model. We suppose the stress intensity factor is 1 under macroscopic scale, and get dimensionless stress intensity factor in the crack tip in two different positions of two materials under relative microscopic scale. We discuss how the stress intensity factor is affected by Nonhomogeneous material under the diverse matrixes and fiber materials, such as elastic modulus ration、volume fraction ratio and Poisson’s ratio.
From the result of this work, we will know no matter the volume fraction, SIFs is smaller than one in fiber when crack tip is in the matrix. When elastic modulus ration raises, the difference increases, too. We observe clearly the change of elastic modulus ration affects slightly SIFs of crack tip in different analytic point of matrix, and concerns strongly in the fiber which is under major external force.

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