Page 88 - Manufacturing Engineering and Technology - Kalpakjian, Serope : Schmid, Steven R.
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Section 2.4  Torsion  67

              Furthermore, friction dissipates energy, so the compressive force is higher than it
              otherwise would be in order to supply the work required to overcome friction. With
              effective lubrication, friction can be minimized, and a reasonably constant cross-
              sectional area can be maintained during the test.
                   When the results of compression and tension tests on ductile metals are com-
              pared, it can be seen that the true stress-true strain curves coincide. This behavior
              does not hold true for brittle materials, which are generally stronger and more duc-
              tile in compression than in tension. (See Table 8.1.)
                   When a metal is subjected to tension into the plastic range, and then the load is
              released and a compressive load is applied, the yield stress in compression is found to
              be lower than that in tension. This phenomenon is known as the Bauschinger effect
              (after ]. Bauschinger, who reported it in 1881), and it is exhibited in varying degrees
              by all metals and alloys. Because of the lowered yield stress in the direction opposite
              that of the original load application, the phenomenon is also called strain softening
              ting (Section 21.2). The test method generally used to
              or work softening.
              determine properties of materials in shear is the torsion
              Disk Test.  For brittle materials such as ceramics and glasses (Chapter 8), a disk test  P
              has been developed, in which the disk is subjected to compression between two hard-
              ened flat platens (Fig. 2.9). When the material is loaded as shown, tensile stresses
              develop perpendicular to the vertical centerline along the disk; fracture begins and
              the disk splits in half vertically. The tensile stress, 0, in the disk is uniform along the
              centerline and can be calculated from the formula                                        Fracture

                                              0' - E,                            (2.10)
                                                   2P

              where P is the load at fracture, cl is the diameter of the disk, and t is its thickness. In  P
              order to avoid premature failure at the contact points, thin strips of soft metal are
              placed between the disk and the platens. These strips also protect the platens from  FIGURE 2.9  Disk test on a
              being damaged during the test. The fracture at the center of the specimen has been  brittle material, showing the
              utilized in the manufacture of seamless tubing (Section 13.6).             direction of loading and the
                                                                                         fracture path.


              2.4    Torsion

              In addition to undergoing tension and compression, a workpiece may be subjected
              to shear strains (Fig. 2.1O), such as in the punching of holes in sheet metals (Sec-
              tion 16.2), in swaging (Section 14.4), and in metal cut-       ->| /|<-                     Q5  r(1)
                   The torsion specimen usually has a reduced cross                                     V
                                                                                   =======
                                                                   ========
              test. In order to obtain an approximately uniform stress   if                             gg'
              and strain distribution along the cross section, this test                      `  `
              is usually performed on a thin tubular specimen.

              section in order to confine the deformation to a narrow                                   t
                                                                                        U
              zone. The shear stress can be calculated from the                        _ ..,§.\  'T _L
              formula                                                                           _F nb
                                 1' = T1                 (2.11)                             lT’l
                                       T
                                     2117215
                                                                  FIGURE 2.l0  A typical torsion-test specimen;  it  is
              where T is the torque,  1' is the average radius of the  mounted between the two heads of a testing machine and
              tube, and t is the thickness of the tube at its narrow  twisted. Note the shear deformation of an element in the
              section.                                           reduced section of the specimen.
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