372 Mechanics of Materials §14.11 During straining the line PN rotates counterclockwise through a small angle a. (exacos 0)cos0-acos20e acos 0 sin0 =(Ex-Ee)cot0 The line OM also rotates,but clockwise,through a small angle B=(xacos+asin)sin-(e,asin)cos Thus the required shear strain ye in the direction OM,i.e.the amount by which the angle OPN changes,is given by Yo=+B=(Ex-Eo)cot+(Ex cos0+Yx>sin0)sin 0-E sin e cos0 Substituting for 8e from egn.(14.14)gives yg=2(ex-e,)cosθsin6-yxy(cos2θ-sin2) Ye=(ex-e)sin 20-1Yxy cos 20 which again is similar in form to the expression for the shear stress t on any inclined plane 0. For consistency of sign convention,however (see $14.11 below),because OM'moves clockwise with respect to OM it is considered to be a negative shear strain,i.e. 生Ya=-[(ex-e,)sin28-生yx,cos20] (14.15) 14.11.Principal strain-Mohr's strain circle Since the equations for stress and strain on oblique planes are identical in form,as noted above,it is evident that Mohr's stress circle construction can be used equally well to represent strain conditions using the horizontal axis for linear strains and the vertical axis for half the shear strain.It should be noted,however,that angles given by Mohr's stress circle refer to the directions of the planes on which the stresses act and not to the direction of the stresses themselves.The directions of the stresses and hence the associated strains are therefore normal (i.e.at 90)to the directions of the planes.Since angles are doubled in Mohr's circle construction it follows therefore that for true similarity of working a relative rotation of the axes of 2 x 90=180 must be introduced.This is achieved by plotting positive shear strains vertically downwards on the strain circle construction as shown in Fig.14.10. Fig.14.10.Mohr's strain circle.372 Mechanics of Materials 614.1 1 During straining the line PN rotates counterclockwise through a small angle u. (&,a cos e) cos e - a cos2 8Ee a cos 8 sin 8 U= = (E, -Ee) cot 8 The line OM also rotates, but clockwise, through a small angle (&,a cos 8 + yxya sin 8) sin 8 - (&,a sin 8) cos 8 a B= Thus the required shear strain Ye in the direction OM, i.e. the amount by which the angle OPN changes, is given by ye = u + B = (E, - Ee)cot 8 + (E,COS 8 + y,,sin 8)sin 8 - &,sin 8 cos 8 Ye = 2 (E, - E,) cos e sin 8 - yxy (cos2 e - sin2 e) Substituting for from eqn. (14.14) gives .. which again is similar in form to the expression for the shear stress ‘t on any inclined plane e. For consistency of sign convention, however (see 6 14.1 1 below), because OM’ moves clockwise with respect to OM it is considered to be a negative shear strain, Le. 370 = 3 (E, - E,) sin 28 -$y,, cos 28 +ye = -[3(~,-&~)~i02e-+y~,~0~28] (14.H) 14.11. Principal strain- Mohr’s strain circle Since the equations for stress and strain on oblique planes are identical in form, as noted above, it is evident that Mohr’s stress circle construction can be used equally well to represent strain conditions using the horizontal axis for linear strains and the vertical axis for halfthe shear strain. It should be noted, however, that angles given by Mohr’s stress circle refer to the directions of the planes on which the stresses act and not to the direction of the stresses themselves. The directions of the stresses and hence the associated strains are therefore normal @e. at 90”) to the directions of the planes. Since angles are doubled in Mohr’s circle construction it follows therefore that for true similarity of working a relative rotation of the axes of 2 x 90 = 180” must be introduced. This is achieved by plotting positive shear strains vertically downwards on the strain circle construction as shown in Fig. 14.10. 0 + +Y Fig. 14.10. Mohr’s strain circle