已知应力状态如图7-15所示，图中应力单位为MPa。试用解析法和图解法求：（1）主应力大小，主平面位置；（2）在单元体上绘出主平面位置及主应力方向：（3）最大切应力。

解析法。由图7-15 (b), 5RBVs9SiqFOgbWM1+33FfjQ5MzVBuwBMtKxl7OJO2aSBrHV8/ZwMOcfoiI5cl0wU+PBHphoKtBlSaOBBP++EDQ==,则GJYfMKjLkofbp2hHpywV2XJKDrhKyY8pNkabQTnZSfNj4EeU04hp7GeRIMqhFLP0LH1/Zor45euqAX+Ib4wy+EW+zi5QNsZwUPdlpvp4G49svGETL7T8P0RELXYT0GSlQnr+VsxZIvVsbtVkL2FNo9q0BUf1A/YTY5p+ZsAo9RjQRzr0LmT9HqsR4zqvqdWxxvwBOOLMpRWubydGX0w1w2i1b6F1y3amlIcP39Xay2W/nBUwvcYhfq1ntdaasVR71M4qSjzMqZGsQ+OXMSBVwZ3caH32tlWLqH+k75tQR4NKZHRhOR8DtzkPy3EbHZZrDepef0rxscHFzW7t6VOysgs4YfSJgd7U5cVw/oOHDSS2pmyO6qejvlmIoelDSCWsL9imY3SPkG4e1hoD+Xmw4v0mHM68xaxzpcf04ez5zEg=因此，主应力为+KScV+y1panwvOoffhg5HhPviuaY4Nkp8eVs+CVvOfto09Dat9iv0uuIv1geR7Zl25M/wDC2NMu3jeyaV7SdpA==,则uHU1LlMA/MDVmXjkTzbge7o5rIe/d+aLpJgd9ot2h2uTNSd0f+9cgIQBgmuN4A63+hrRm23NxnyQzBYwGGEjXAbgprbzsStlYXHS17PMVlobgTInTcPPFFqQfQaZNejbO1Rz4JoJlWXzmAWlJx2DiO7J8on+6pouIYsU/leWza4= 或aWPnpQq+gHaV3AO3j4o0KdmRHJ4SaUQnkbH8y2hVn0w=因为foSbrp1iWk/2N47CuJoQgldFesVuEX79705H9/kJQpc=，所以两个方向角中，绝对值较小的一个，即XwDkf41CweJ4p/XI+7OIUpCXB7TmKNJIrg04QBcSK0g=为正应力。所在的主平面，主单元体方位如图7—17 （a）所示。最大切应力为3pDUMInTFlAeiueFSZOpU4FaIkEj3Hgq/6ldpHPtQ44UJzIGFxEnHhy2jyX8W2S8nRcw2riASrMvoila2TwjIKwC5ojAAvbGKlm/yDk4KjA=17d1defd7aecf66.png图解法。首先，根据图7—15 （b）所示单元体上的应力，确定应力圆上的两个点4od7Id5c94U14iRtpXqTfIVZd2nQMQuNs+kddV1EDjEspcgOlp4LHbLvT5UflKB3和 JsznLJPvIz55xucM+eEXR42bPSvONp+3L/mn/gNkHJaIc5CHclB8ph9ECDCpjXJ7,作出应力圆如图7-17 (b)所示, 由应力圆与。轴的交点得到wVs3w9YRTdhIkyly0cDyHS5vw9sYREmHgDqrk66+3mypnGAL1KVW6M0a902QHBZRhGyIWhez8hy8oMb2vsb39Q==，因此，主应力8c/dWter8k7EI+h+Wllx4A0/skmiueH5RZ4xz+QqjXK0dE0Vn5+1Z8HdaBSFHUA/HU5n5HfsZNpXMsB9LA6W5w==，应力圆上点mzzTehuQMrP8CQIGBwwVw8pgbK6omg6sgdiSHtEpWSA=所对应的半径与 轴正向的夹角为主方向的二倍角 riAEWhxfpug5DDOSkFQT7PBYznZAlxcgogC3VgiF3SPBzTV9YWoYc1rEmOlvpGz4, 如图 7-17 (b) 所示, 由此得到 pDJJ8ZGA7voK320WdFFXYoQONKTlgi5oP9v8PwKtjRI=应力圆上纵 坐标的最大值为最大切应力. 由图 7-17 (b) 得到 qWxtrNNQOb3LGibpVIccfY/vXozdwlc7uX20r249SkqrAdj9O9zbNPWpj3oHVFmO