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10  Basic elasticity

             where I, m and n become the direction cosines of  the  angles that  a normal  to the
             surface of the body makes with the x, y  and z axes respectively.


                1.6  Determination of stresses on inclined planes

             The complex stress system of Fig.  1.6 is derived from a consideration  of the actual
             loads  applied  to  a  body  and  is  referred  to  a  predetermined,  though  arbitrary,
             system of axes. The values of these stresses may not give a true picture of the severity
             of stress at that point so that it is necessary to investigate the state of stress on other
             planes on which the direct and shear stresses may be greater.
               We shall restrict the analysis to the two-dimensional system of plane stress defined
             in Section 1.4.
               Figure  1.8(a) shows a complex stress system at a point in a body referred to axes
             Ox, Oy. All  stresses are positive as defined in  Section  1.2. The shear  stresses T~~
             and rYx were shown to be equal in Section 1.3. We now, therefore, designate them
             both  rxY. The element of  side Sx, Sy  and  of  unit  thickness is  small so that  stress
             distributions over  the  sides of  the  element may  be  assumed to be  uniform.  Body
             forces are ignored since their contribution is a second-order term.
               Suppose that we require to find the state of stress on a plane AB inclined at an angle
             8 to the vertical. The triangular  element EDC formed by the plane and the vertical
             through  E is  in  equilibrium under  the action  of  the  stresses shown in  Fig.  1.8(b),
             where a, and  T are the direct and shear components of the resultant  stress on AB.
             Then resolving forces in a direction perpendicular to ED we have
                       a,ED  = a,EC  cos 8 + a,CD  sin 8 + T,,EC  sin 8 + T,,CD  cos 8
             Dividing through by ED and simplifying
                                  a, = ax cos2 8 + aY sin2 8 + T~, sin 28
               Now resolving forces parallel to ED
                        TED = a,EC  sin 8 - ayCD cos 8 - T,,EC  cos 8 + T,,CD  sin 8




                                                                        E











              0

                               (a)                                        (b)
              Fig. 1.8  (a) Stresses on a two-dimensional element; (b) stresses on an inclined plane at the point.
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