The Physical Nature of Materials Strengths

Abstract

The strength of a material is assessed most often by means of a tensile test. For a given material with an original cross-section area A 0 , if the applied maximum tensile force is equal to F max , the fracture strength can be calculated by r F =F max /A 0 , as described in the textbooks. [1,2] For a bulk metallic glassy specimen, it often fails in a shear mode, as shown in Figure 1, and the shear fracture surface makes an angle of h T =56°with respect to the tension axis. Such shear fracture behavior has been widely observed in many metallic glasses, as summarized in the literatures [3,4] and Table 1. [5][6][7][8][9][10][11][12][13][14] According to the definition in the textbooks, [1,2] the tensile fracture strength of the metallic glass should be equal to r T =F max /A 0 . However, the actual area of the shear fracture surface becomes A 0 /sin(h T ) and the applied normal tensile force on the shear plane is F max cos(h T ), as shown in Figure 1. This will result in another tensile strength F max sin(h T )cos(h T )/ A 0 , which is different from that (F max /A 0 ) defined in textbooks. [1,2] Consequently, this gives rise to some interesting and significant questions. Which is the real tensile strength of a metallic glass, F max /A 0 or F max sin(h T )cos(h T )/A 0 ? Why do metallic glasses often fail neither along the maximum normal stress plane (h T =90°) nor along the maximum shear stress plane (h T =45°) under tensile loading [5][6][7][8][9][10][11][12][13][14] ? What is the physical nature of the materials strength?For a material subjected to a tensile force F, there is always a combined stress (r n ,s n ) on any plane, as illustrated in Figure 2(a). In order to better understand the physical nature of materials strength and answer the interesting questions above, we proposed that there are only two independent intrinsic strengths r 0 and s 0 for an isotropic material 4 . As illustrated in Figure 2(b), r 0 is defined as the critical strength of a material in a Mode I failure; s 0 is the critical strength of a material in a Mode II fracture. If any plane of a material is subjected to a combined stress (r n ,s n ), the tensile failure condition can be expressed by the following criterion 4 , (r n /r 0 ) 2 + (s n /s 0 ) 2 = 1.(1)Meanwhile, the tensile stress state (r n ,s n ) on any shear plane follows the Mohr-circle equation, i.e.(r n -r T /2) 2 + (s n ) 2 = (r T /2) 2 .(2)According to Equations 1 and 2, the two independent strengths, s 0 and r 0 can be derived as:Here, r T =F max /A 0 is the so-called tensile fracture strength; a is the fracture mode factor [4] and can be calculated from the macroscopic tensile shear fracture angle, see the discussion below. Here, it is suggested that r T =F max /A 0 can be only regarded as nominal fracture strength, rather than the intrin-COMMUNICATIONS ADVANCED ENGINEERING MATERIALS 2007, 9, No. 3 metallic glassy specimen. The nominal tensile fracture strength of the specimen is about 1.58 GPa and the shear fracture surface makes an angle of about 56°with respect to the t...

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