. The strength of materials; a text-book for engineers and architects. the previous case. In the limiting case in which the direct stress / is zero we 5s .get p ^ -, a stress shear is equivalent to a normal stress of four-fifths of the shear stress, or the shear strength of a materialis four-fifths of the tensile strength. 3. Equivalent Shear Stress or Guest or TrescaTheory.—According to this theory, which is associated withthe name of Mr. J. J. Guest, who was one of the first tocarry out careful experiments upon the subject, and is usuallyattributed to Tresca, failure occurs by sliding o
. The strength of materials; a text-book for engineers and architects. the previous case. In the limiting case in which the direct stress / is zero we 5s .get p ^ -, a stress shear is equivalent to a normal stress of four-fifths of the shear stress, or the shear strength of a materialis four-fifths of the tensile strength. 3. Equivalent Shear Stress or Guest or TrescaTheory.—According to this theory, which is associated withthe name of Mr. J. J. Guest, who was one of the first tocarry out careful experiments upon the subject, and is usuallyattributed to Tresca, failure occurs by sliding of the particlesover each other, i. e. by shear. From p. 20 we get Equivalent shear stress = . +52 ^4^ and acts at an angle of 45° to the normal stress. BEHAVIOUR OF MATERIALS UNDER TEST 45 To compare a simple shear with a simple tension orcompression by this formula we put s = o and we get Equivalent shear stress = ~ i. e. a shear stress is equivalent to a normal stress of twice itsmagnitude or the shear strength of a material is one-half of itstensile Fig. 19.—Naviers Theory. It is interestmg to note that as shown on p. 32 theequivalent shear stress comes the same whether worked fromthe point of view of stress or of strain, and so there is logicalsupport for this theory. 4. Sliding with Internal Friction or Navier Theory.—This theory deals with materials subjected to compressivestresses and attempts to explain the fact that short cylinders 46 THE STRENGTH OF MATERIALS of brittle material usually fail by sliding or shearing alonga line e f, Fig. 19, inclined at an angle 0 from 55° to 65°, onthe ground that the particles are capable of exerting frictionalresistances. Consider a short column a b c D of unit sectional areasubjected to an ultimate compressive force u,, which causesfailure, and consider the forces across a section E f, theultimate or breaking shear stress in the material being u,. The force u,. acting along the section e f can be resolvedinto she
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