328 Mechanics of Materials S13.2 A y Fig.13.2.Stresses on an element subjected to pure shear. Consider now the equilibrium of portion PBC. Resolving normal to PC assuming unit depth, dox PC=txy×BC sin0+txy×PB cos0 =txy×PC cos sinθ+txy×PC sin0 cos0 o=txy sin 20 (13.3) The maximum value of ce is txy when 6=45. Similarly,resolving forces parallel to PC, tox PC txy x PB sin 0-txy BC cos0 =txw×PC sin20-ty×PCcos20 to=-txy Cos 20 (13.4) The negative sign means that the sense of t is opposite to that assumed in Fig.13.2. The maximum value of te is xy when=0or 90and it has a value of zero when=45, i.e.on the planes of maximum direct stress. Further consideration ofegns.(13.3)and(13.4)shows that the system of pure shear stresses produces an equivalent direct stress system as shown in Fig.13.3,one set compressive and one tensile,each at 45 to the original shear directions,and equal in magnitude to the applied shear. G-Twy 451 0=--Txy Fig.13.3.Direct stresses due to shear. This has great significance in the measurement of shear stresses or torques on shafts using strain gauges where the gauges are arranged to record the direct strains at 45 to the shaft axis. Practical evidence of the theory is also provided by the failure of brittle materials in shear. A shaft of a brittle material subjected to torsion will fail under direct stress on planes at 45to the shaft axis.(This can be demonstrated easily by twisting a piece of blackboard chalk in328 Mechanics of Materials $13.2 Fig. 13.2. Stresses on an element subjected to pure shear. Consider now the equilibrium of portion PBC. Resolving normal to PC assuming unit depth, no x PC = zxy x BC sin8+zxy x PB cos8 (13.3) = zxy x PC cos 8 sin 8 + zxy x PC sin 8 cos 8 .. The maximum value of no is zxy when 8 = 45". Similarly, resolving forces parallel to PC, q, = zXy sin 28 z, x PC = zxy x PB sin 8 - zxy BC cos 8 = zxy x PC sin2 8 - zXy x PC cos2 e .. -zxycos28 (13.4) The negative sign means that the sense of To is opposite to that assumed in Fig. 13.2. The maximum value of 7, is zxy when 8 = 0" or 90" and it has a value of zero when 8 = 45", i.e. on the planes of maximum direct stress. Further consideration of eqns. (13.3) and (13.4) shows that the system of pure shear stresses produces an equivalent direct stress system as shown in Fig. 13.3, one set compressive and one tensile, each at 45" to the original shear directions, and equal in magnitude to the applied shear. XY Fig. 13.3. Direct stresses due to shear. This has great sign@ance in the measurement of shear stresses or torques on shafts using strain gauges where the gauges are arranged to record the direct strains at 45" to the shaft axis. Practical evidence of the theory is also provided by the failure of brittle materials in shear. A shaft of a brittle material subjected to torsion will fail under direct stress on planes at 45" to the shaft axis. (This can be demonstrated easily by twisting a piece of blackboard chalk in