§8.4 Introduction to Advanced Elasticity Theory 225 Pn*resultont stress on ABC n normol to ABC n 只0 Cartesian components of resultont 5tre多s Fig.8.5.Cartesian components of resultant stress on an inclined plane. Fig.8.6.Normal and tangential components of resultant stress on an inclined plane. or p听=p层n+p+p (8.17) these being alternative forms of eqns.(8.6)and(8.4)respectively. From eqn.(8.5)the normal stress on the plane is given by: On=pxn·【+Pw·m+Pn·n But from eqns.(8.13),(8.14)and(8.15) Pxt=Oxx·l十Oxy·m+O·n pm=ox·l+oy·m+ox·n pn=ox1+oy·m+ox·n .Substituting into eqn (8.5)and using the relationships oxy=ox:ox=ox and o=o which will be proved in $8.12 om=aa·l2+ow·m2+az·n2+2ay·lm+2oz·mn+2oz·ln. (8.18)
$8.4 Introduction to Advanced Elasticity Theory 225 t' X P, sresultont stress n=normol to ABC on ABC P, Cortesion I Components pv ot resultant p,, 1 stress Y Fig. 8.5. Cartesian components of resultant stress on an inclined plane X Fig. 8.6. Normal and tangential components of resultant stress on an inclined plane. 2 or ~,2 =P:n + P;n + PW these being alternative forms of eqns. (8.6) and (8.4) respectively. From eqn. (8.5) the normal stress on the plane is given by: (8.17) an = Pxn .I + pbII . m + pill . But from eqns. (8.13), (8.14) and (8.15) pxr, = a,, . I + a,\ . m + a,,? . n Pyn = aVx .1 + a), . m + 0): . n pzn = a, . I + a,, . m + az, . n :. Substituting into eqn (8.5) and using the relationships a,, = ayx; ax, = which will be proved in 58.12 and a,: = a,, Un =uxx~12+uyy~m 2 +u,.n 2 +2u,,~lm+2uy~~mn+2ux~~ln. (8.18)
226 Mechanics of Materials 2 s8.4 and from egn.(8.6)the shear stress on the plane will be given by 六=p层n+pn+p味-房 (8.19) In the particular case where plane ABC is a principal plane(i.e.no shear stress): 0xy=0x=0z=0 and 0xx=o1,0y=02and0x=03 the above equations reduce to: gn=o112+02·m2+3·n2 (8.20) and since pxm=o·Lpym=o2·m and p2n=o3·n 员=2+吃m2+n2- (8.21) 8.4.1.Line of action of resultant stress As stated above,the resultant stress p is generally not normal to the plane ABC but inclined to the x,y and z axes at angles 6x,6 and 0:-see Fig.8.7. Fig.8.7.Line of action of resultant stress. The components of p in the x,y and z directions are then Pxn=Pm.Cos6x） Pyn Pn.cos (8.22) Pan Pn.cos.) and the direction cosines which define the line of actions of the resultant stress are 1'=cos0x pxn/Pn m'=cos0y Pyn/pn (8.23) n'cos0:P:n/Pn)
226 Mechanics of Materials 2 $8.4 and from eqn. (8.6) the shear stress on the plane will be given by 4, =P:n +P;n +Pzn - 02, (8.19) In the particular case where plane ABC is a principal plane (i.e. no shear stress): and the above equations reduce to: 8.4.1. Line of action of resultant stress As stated above, the resultant stress p,* is generally not normal to the plane ABC but inclined to the x, y and z axes at angles ex, 8" and 6, - see Fig. 8.7. i Fig. 8.7. Line of action of resultant stress. The components of pn in the x, y and z directions are then 1 pxn = Pn . COS 0, Pyn = Pn. COS 0.v P2n = pn. cos ez and the direction cosines which define the line of actions of the resultant stress are 1' = COS 0.r = P.m l Pn m' = cos0, = pynn/pn n' = cos OZ = p_,* /P,~ (8.22) (8.23)
§8.4 Introduction to Advanced Elasticity Theory 227 8.4.2.Line of action of normal stress By definition the normal stress is that which acts normal to the plane,i.e.the line of action of the normal stress has the same direction cosines as the normal to plane viz:I,m and n. 8.4.3.Line of action of shear stress As shown in $8.4 the resultant stress pn can be considered to have two components;one normal to the plane (o)and one along the plane (the shear stress )-see Fig.8.6. Let the direction cosines of the line of action of this shear stress be ls,ins and ns. The alternative components of the resultant stress,Px,Pym and pn,can then either be obtained from eqn(8.22)or by resolution of the normal and shear components along the x, y and z directions as follows: Pxn=on·l+tn·lg pyn=oa·m十tn·ms (8.24) P2n=on·n+tn·ns Thus the direction cosines of the line of action of the shear stress t are: l=Pxn-1.on Tn ms= Pyn-m·gn (8.25) In %, Pn-i·on En 8.4.4.Shear stress in any other direction on the plane Let be the angle between the direction of the shear stress tm and the required direction. Then,since the angle between any two lines in space is given by, cosφ=l,·lΦ+ms·mo+ns·no (8.26) where lo,mo,n are the direction cosines of the new shear stress direction,it follows that the required magnitude of the shear stress on the"o"plane will be given by o=tm·C0S中 (8.27) Alternatively,resolving the components of the resultant stress (pn,Py and pan)along the new direction we have: to=Pxm·l6+pw·mo+Pm·no (8.28) and substituting egns.(8.13),(8.14)and (8.15) to=oxx·lw十oy·mme+oz·nno+oy(mΦ+lo·m) Oxz (ino nlo)+oyz (mno nmo) (8.29) Whilst egn.(8.28)has been derived for the shear stress to it will,in fact,apply equally for any type of stress (i.e.shear or normal)which acts on the plane ABC in the direction. In the case of the shear stress,however,its line of action must always be perpendicular to the normal to the plane so that llo+mm中+nnΦ=0
58.4 Introduction to Advanced Elasticiiy Theory 227 8.4.2. Line of action of normal stress By definition the normal stress is that which acts normal to the plane, i.e. the line of action of the normal stress has the same direction cosines as the normal to plane viz: 1, m and n. 8.4.3. Line of action of shear stress As shown in 58.4 the resultant stress p,, can be considered to have two components; one normal to the plane (a,,) and one along the plane (the shear stress t,,) - see Fig. 8.6. Let the direction cosines of the line of action of this shear stress be l,, in, and n,. The alternative components of the resultant stress, pxn, py,, and pzn, can then either be obtained from eqn (8.22) or by resolution of the normal and shear components along the x, y and z directions as follows: (8.24) pxn = an -1 + tn 1s Pyn = an * m + rn * m, Pzn = an * n + rn ns Thus the direction cosines of the line of action of the shear stress t,, are: 1 Pxn - 1 an fn Pyn - m ‘ an rn Pzn - n an rn 1, = m, = n, = (8.25) 8.4.4. Shear stress in any other direction on the plane Let 4 be the angle between the direction of the shear stress r,, and the required direction. cos4 = 1,. 14 + m, . rn4 + n, . n+ (8.26) where 14, m4, n4 are the direction cosines of the new shear stress direction, it follows that the required magnitude of the shear stress on the “q5” plane will be given by z+ = T, * cos4 (8.27) Alternatively, resolving the components of the resultant stress (pXlt, pvn and pz,,) along the new direction we have: (8.28) Then, since the angle between any two lines in space is given by, = Pxn + Pyfl . “4 + Pzn ’ ng and substituting eqns. (8.13), (8.14) and (8.15) q = a,, 11+ + uyy - mmg + a, - nn+ + axy (lm+ + 1, - m 1 + axz (In+ + nl+) + ayz (mn+ + nm+) (8.29) Whilst eqn. (8.28) has been derived for the shear stress tg it will, in fact, apply equally for any type of stress (Le. shear or normal) which acts on the plane ABC in the 4 direction. In the case of the shear stress, however, its line of action must always be perpendicular to the normal to the plane so that llg + rnm4 + nng = 0
228 Mechanics of Materials 2 $8.5 In the case of a normal stress the relationship between the direction cosines is simply l=l,m=mo and n=n中 since the stress and the normal to the plane are in the same direction.Egn.(8.29)then reduces to that found previously,viz.eqn.(8.18). 8.5.Principal stresses and strains in three dimensions-Mohr's circle representation The procedure used for constructing Mohr's circle representation for a three-dimensional principal stress system has previously been introduced in $13.77.For convenience of refer- ence the resulting diagram is repeated here as Fig.8.8.A similar representation for a three-dimensional principal strain system is shown in Fig.8.9. T Principal circle 2 Fig.8.8.Mohr circle representation of three-dimensional stress state showing the principal circle,the radius of which is equal to the greatest shear stress present in the system. Principal arcle Fig.8.9.Mohr representation for a three-dimensional principal strain system. EJ.Heam.Mechanics of Materials 1.Butterworth-Heinemann.1977
228 Mechanics of Materials 2 $8.5 In the case of a normal stress the relationship between the direction cosines is simply 1 = 14, m = m# and n = n6 since the stress and the normal to the plane are in the same direction. Eqn. (8.29) then reduces to that found previously, viz. eqn. (8.18). 8.5. Principal stresses and strains in three dimensions - Mohr’s circle representation The procedure used for constructing Mohr’s circle representation for a three-dimensional principal stress system has previously been introduced in Q 13.7? For convenience of reference the resulting diagram is repeated here as Fig. 8.8. A similar representation for a three-dimensional principal strain system is shown in Fig. 8.9. t Fig. 8.8. Mohr circle representation of three-dimensional stress state showing the principal circle, the radius of which is equal to the greatest shear stress present in the system. Fig. 8.9. Mohr representation for a three-dimensional principal strain system. J. ’ E.J. Hearn. Mechanics of Materids I, Butterworth-Heinemann, 1977
§8.6 Introduction to Advanced Elasticity Theory 229 In both cases the principal circle is indicated,the radius of which gives the maximum shear stress and half the maximum shear strain,respectively,in the three-dimensional system. This form of representation utilises different diagrams for the stress and strain systems. An alternative procedure uses a single combined diagram for both cases and this is described in detail $$8.6 and 8.7. 8.6.Graphical determination of the direction of the shear stress on an inclined plane in a three-dimensional principal stress system As before,let the inclined plane have direction cosines l,m and n.A true representation of this plane is given by constructing a so-called "true shape triangle"the ratio of the lengths of its sides being the ratio of the direction cosines-Fig.8.10. Fig.8.10.Graphical determination of direction of shear stress on an inclined plane. If lines are drawn perpendicular to each side from the opposite vertex,meeting the sides at points P,R and S,they will intersect at point T the"orthocentre".This is also the point through which the normal to the plane from O passes. If 2 and o3 are the three principal stresses then point M is positioned on AC such that CM_(a2-3】 CA(o1-02) The required direction of the shear stress is then perpendicular to the line BD. The equivalent procedure on the Mohr circle construction is as follows (see Fig.8.1 1). Construct the three stress circles corresponding to the three principal stresses o:,o2 and o3. Set off line AB at an angle a cos-I to the left of the vertical through A. Set off line CB at an angle y=cos-!n to the right of the vertical through C to meet AB at B. Mark the points where these lines cut the principal circle R and P respectively Join AP and CR to cut at point T. Join BT and extend to cut horizontal axis AC at S. With point M the o2 position,join BM. The required shear stress direction is then perpendicular to the line BM
98.6 Introduction to Advanced Elasticity Theory 229 In both cases the principal circle is indicated, the radius of which gives the maximum shear stress and hay the maximum shear strain, respectively, in the three-dimensional system. This form of representation utilises different diagrams for the stress and strain systems. An alternative procedure uses a single combined diagram for both cases and this is described in detail $98.6 and 8.7. 8.6. Graphical determination of the direction of the shear stress tn on an inclined plane in a three-dimensional principal stress system As before, let the inclined plane have direction cosines 1, m and n . A true representation of this plane is given by constructing a so-called “true shape triangle” the ratio of the lengths of its sides being the ratio of the direction cosines-Fig. 8.10. Fig. 8.10. Graphical determination of direction of shear stress on an inclined plane. If lines are drawn perpendicular to each side from the opposite vertex, meeting the sides at points P, R and S, they will intersect at point T the “orthocentre”. This is also the point through which the normal to the plane from 0 passes. If 01,02 and a3 are the three principal stresses then point M is positioned on AC such that CM (a2 - 03) CA (01 - 02) The equivalent procedure on the Mohr circle construction is as follows (see Fig. 8.1 1). -- - The required direction of the shear stress is then perpendicular to the line BD. Construct the three stress circles corresponding to the three principal stresses a1 ,02 and 03. Set off line AB at an angle a! = cos-’ 1 to the left of the vertical through A, Set off line CB at an angle y = cos-’ n to the right of the vertical through C to meet AB at B. Mark the points where these lines cut the principal circle R and P respectively. Join AP and CR to cut at point T. Join BT and extend to cut horizontal axis AC at S. With point M the 02 position, join BM. The required shear stress direction is then perpendicular to the line BM