§13.7 Complex Stresses 335 28 G+) 6g=克(6，+)+主(c-6)o:26 T6=克(-s:)n28 Fig.13.11.Free-hand sketch of Mohr's stress circle. Equations(13.8)and(13.9)give the values of the direct stress and shear stress te on any plane inclined at an angle e to the plane on which the direct stress ox acts within a two- dimensional complex stress system,viz: o=(ax+)+(ax-0)cos 20+txy sin 20 te =(ax-0)sin 20-txy cos 20 (a)Uniaxial stresses For the special case of a single uniaxial stress o,as in simple tension or on the surface of a beam in bending,o,=txy =0 and the equations (13.8)and (13.9)reduce to 0g=0x(1+cos20)=0xcos20. N.B.If the single stress were selected as o,then the relationship would have reduced to that of eqn.(13.1),i.e. e=ay sin20. Similarly: to=ax sin 20. Plotting these equations on simple Cartesian axes produces the stress distribution diagrams of Fig.13.12,both sinusoidal in shape with shear stress"shifted"by 45 from the normal stress. Principal stresses op and occur,as expected,at 90 intervals and the amplitude of the normal stress curve is given by the difference between the principal stress values.It should also be noted that shear stress is proportional to the derivative of the normal stress with respect to 0,i.e.te is a maximum where doe/do is a maximum and te is zero where doe/de is zero,etc. Alternatively,plotting the same equations on polar graph paper,as in Fig.13.13,gives an even more readily understood pictorial representation of the stress distributions showing a peak value of direct stress in the direction of application of the applied stressox falling to zero$13.7 Complex Stresses tt 335 '6 Fig. 13.11. Free-hand sketch of Mohr's stress circle. Equations (13.8) and (13.9) give the values of the direct stress ug and shear stress re on any plane inclined at an angle 8 to the plane on which the direct stress u, acts within a two￾dimensional complex stress system, viz: ug = *(a, + a,) +3(ux - 0,) cos 28 + s,, sin 28 q, = i(u, - cy) sin 28 - T~~ cos 28 (a) Uniaxial stresses For the special case of a single uniaxial stress ux as in simple tension or on the surface of a beam in bending, u, = zxy = 0 and the equations (13.8) and (13.9) reduce to 00 = $0, (1 + COS 28) = 6, COS' 8. N.B. If the single stress were selected as u, then the relationship would have reduced to that of eqn. (13.1), i.e. ae = uy sin' 8. Similarly: 70 = 3 ox sin 28. Plotting these equations on simple Cartesian axes produces the stress distribution diagrams of Fig. 13.12, both sinusoidal in shape with shear stress "shifted by 45" from the normal stress. Principal stresses op and oq occur, as expected, at 90" intervals and the amplitude of the normal stress curve is given by the difference between the principal stress values. It should also be noted that shear stress is proportional to the derivative of the normal stress with respect to 8, i.e. 70 is a maximum where doe/d8 is a maximum and 70 is zero where da,/d8 is zero, etc. Alternatively, plotting the same equations on polar graph paper, as in Fig. 13.13, gives an even more readily understood pictorial representation of the stress distributions showing a peak value of direct stress in the direction of application of the applied stress ox falling to zero