CHAPTER 8 INTRODUCTION TO ADVANCED ELASTICITY THEORY 8.1.Type of stress Any element of material may be subjected to three independent types of stress.Two of these have been considered in detai previously,namely direct stresses and shear stresses, and need not be considered further here.The third type,however,has not been specifically mentioned previously although it has in fact been present in some of the loading cases considered in earlier chapters;these are the so-called body-force stresses.These body forces arise by virtue of the bulk of the material,typical examples being: (a)gravitational force due to a component's own weight:this has particular significance in civil engineering applications,e.g.dam and chimney design; (b)centrifugal force,depending on radius and speed of rotation,with particular significance in high-speed engine or turbine design; (c)magnetic field forces. In many practical engineering applications the only body force present is the gravitational one,and in the majority of cases its effect is minimal compared with the other applied forces due to mechanical loading.In such cases it is therefore normally neglected.In high- speed dynamic loading situations such as the instances quoted in (b)above,however,the centrifugal forces far exceed any other form of loading and are therefore the primary factor for consideration. Unlike direct and shear stresses,body force stresses are defined as force per unit volume, and particular note must be taken of this definition in relation to the proofs of formulae which follow. 8.2.The cartesian stress components:notation and sign convention Consider an element of material subjected to a complex stress system in three dimensions. Whatever the type of applied loading the resulting stresses can always be reduced to the nine components,i.e.three direct and six shear,shown in Fig.8.1. It will be observed that in this case a modified notation is used for the stresses.This is termed the double-suffix notation and it is particularly useful in the detailed study of stress problems since it indicates both the direction of the stress and the plane on which it acts. The first suffix gives the direction of the stress. The second suffix gives the direction of the normal of the plane on which the stress acts. Thus,for example, 220
CHAPTER 8 INTRODUCTION TO ADVANCED ELASTICITY THEORY 8.1. Type of stress Any element of material may be subjected to three independent types of stress. Two of these have been considered in detai previously, namely direct stresses and shear stresses, and need not be considered further here. The third type, however, has not been specifically mentioned previously although it has in fact been present in some of the loading cases considered in earlier chapters; these are the so-called body-force stresses. These body forces arise by virtue of the bulk of the material, typical examples being: (a) gravitational force due to a component’s own weight: this has particular significance in (b) centrifugal force, depending on radius and speed of rotation, with particular significance (c) magnetic field forces. civil engineering applications, e.g. dam and chimney design; in high-speed engine or turbine design; In many practical engineering applications the only body force present is the gravitational one, and in the majority of cases its effect is minimal compared with the other applied forces due to mechanical loading. In such cases it is therefore normally neglected. In highspeed dynamic loading situations such as the instances quoted in (b) above, however, the centrifugal forces far exceed any other form of loading and are therefore the primary factor for consideration. Unlike direct and shear stresses, body force stresses are defined as force per unit volume, and particular note must be taken of this definition in relation to the proofs of formulae which follow. 8.2. The Cartesian stress components: notation and sign convention Consider an element of material subjected to a complex stress system in three dimensions. Whatever the type of applied loading the resulting stresses can always be reduced to the nine components, i.e. three direct and six shear, shown in Fig. 8.1. It will be observed that in this case a modified notation is used for the stresses. This is termed the double-suffix notation and it is particularly useful in the detailed study of stress problems since it indicates both the direction of the stress and the plane on which it acts. Thefirst suffix gives the direction of the stress. The second suffix gives the direction of the normal of the plane on which the stress acts. Thus, for example, 220
$8.3 Introduction to Advanced Elasticity Theory 221 0码 Fig.8.1.The cartesian stress components. or is the stress in the X direction on the X facing face (i.e.a direct stress).Common suffices therefore always indicate that the stress is a direct stress.Similarly,ory is the stress in the X direction on the Y facing face (i.e.a shear stress).Mixed suffices always indicate the presence of shear stresses and thus allow the alternative symbols osy or tx.Indeed,the alternative symbol t is not strictly necessary now since the suffices indicate whether the stress o is a direct one or a shear. 8.2.1.Sign conventions (a)Direct stresses.As always,direct stresses are assumed positive when tensile and nega- tive when compressive. (b)Shear stresses.Shear stresses are taken to be positive if they act in a positive cartesian (X.Y or Z)direction whilst acting on a plane whose outer normal points also in a positive cartesian direction. Thus positive shear is assumed with direction and facing face. Alternatively',positive shear is also given with-direction and-facing face (a double negative making a positive,as usual). A careful study of Fig.8.1 will now reveal that all stresses shown are positive in nature. The cartesian stress components considered here relate to the three mutually perpendicular axes X,Y and Z.In certain loading cases,notably those involving axial symmetry,this system of components is inconvenient and an alternative set known as cylindrical components is used.These involve the variables,radius r.angle 6 and axial distance z.and will be considered in detail later. 8.3.The state of stress at a point Consider any point O within a stressed material,the nine cartesian stress components at O being known.It is now possible to determine the normal,direct and resultant stresses which act on any plane through O whatever its inclination relative to the cartesian axes. Suppose one such plane ABC has a normal n which makes angles nx.ny and nz with the YZ.XZ and XY planes respectively as shown in Figs.8.2 and 8.3.(Angles between planes
58.3 Introduction to Advanced Elasticity Theory -% Y 22 1 Fig. 8.1. The Cartesian stress components. a,, is the stress in the X direction on the X facing face (i.e. a direct stress). Common suffices therefore always indicate that the stress is a direct stress. Similarly, a,, is the stress in the X direction on the Y facing face (i.e. a shear stress). Mixed suffices always indicate the presence of shear stresses and thus allow the alternative symbols oxy or xx,. Indeed, the alternative symbol x is not strictly necessary now since the suffices indicate whether the stress a is a direct one or a shear. 8.2.1. Sign conventions (a) Direct stresses. As always, direct stresses are assumed positive when tensile and negative when compressive. (b) Shear stresses. Shear stresses are taken to be positive if they act in a positive Cartesian (X, Y or Z) direction whilst acting on a plane whose outer normal points also in a positive Cartesian direction. Thus positive shear is assumed with + direction and + facing face. Alternatively’, positive shear is also given with -- direction and - facing face (a double A careful study of Fig. 8.1 will now reveal that all stresses shown are positive in nature. The cartesian stress components considered here relate to the three mutually perpendicular axes X, Y and Z. In certain loading cases, notably those involving axial symmetry, this system of components is inconvenient and an alternative set known as cylindrical components is used. These involve the variables, radius r, angle 6 and axial distance z, and will be considered in detail later. negative making a positive, as usual). 83. The state of stress at a point Consider any point Q within a stressed material, the nine Cartesian stress components at Q being known. It is now possible to determine the normal, direct and resultant stresses which act on any plane through Q whatever its inclination relative to the Cartesian axes. Suppose one such plane ABC has a normal n which makes angles nx, ny and nz with the YZ, XZ and XY planes respectively as shown in Figs. 8.2 and 8.3. (Angles between planes
222 Mechanics of Materials 2 $8.3 B asn关 Bany y=nz Fig.8.2. A Z (normol to ABC) an B Body-force stresses Surface area ABC.△s introduced at centroid (shown removed from elemert for C simplification】 Fig.8.3.The state of stress on an inclined plane through any given point in a three-dimensional cartesian stress system. ABC and YZ are given by the angle between the normals to both planes n and x,etc.) For convenience,let the plane ABC initially be some perpendicular distance h from so that the cartesian stress components actually acting at can be shown on the sides of the tetrahdedron element ABCO so formed(Fig.8.3).In the derivation below the value of h will be reduced to zero so that the equations obtained will relate to the condition when ABC passes through O. In addition to the cartesian components,the unknown components of the stress on the plane ABC,i.e.Pn,Py and pin,are also indicated,as are the body-force field stress components which act at the centre of gravity of the tetrahedron.(To improve clarity of the diagram they are shown displaced from the element.) Since body-force stresses are defined as forces/unit volume,the components in the X,Y and Z directions are of the form F×△S年 where ASh/3 is the volume of the tetrahedron.If the area of the surface ABC,i.e.AS,is assumed small then all stresses can be taken to be uniform and the component of force in
222 Mechanics of Materials 2 $8.3 I f 7.1-12 Fig. 8.2 A 2 t ;2 I FX Body- farce stresses introduced at centroid (shown rernced frwn element for simpificatm) L AK2 AS Fig. 8.3. The state of stress bn an inclined plane through any given point in a three-dimensional Cartesian stress system. ABC and YZ are given by the angle between the normals to both planes n and x, etc.) For convenience, let the plane ABC initially be some perpendicular distance h from Q so that the Cartesian stress components actually acting at Q can be shown on the sides of the tetrahdedron element ABCQ so formed (Fig. 8.3). In the derivation below the value of h will be reduced to zero so that the equations obtained will relate to the condition when ABC passes through Q. In addition to the Cartesian components, the unknown components of the stress on the plane ABC, i.e. pxn, pvn and pzn, are also indicated, as are the body-force field stress components which act at the centre of gravity of the tetrahedron. (To improve clarity of the diagram they are shown displaced from the element.) Since body-force stresses are defined as forces/unit volume, the components in the X, Y and Z directions are of the form F x AS; where Ash13 is the volume of the tetrahedron. If the area of the surface ABC, i.e. AS, is assumed small then all stresses can be taken to be uniform and the component of force in
§8.3 Introduction to Advanced Elasticity Theory 223 the X direction due to o is given by oa△S cosnx Stress components in the other axial directions will be similar in form. Thus,for equilibrium of forces in the X direction, PiAS+F,AS=u AS cosnx+tyAS cosny+iAS cosnz As h0(i.e.plane ABC passes through o),the second term above becomes very small and can be neglected.The above equation then reduces to Pxn =Oxx cos nx txy cos ny txz cos nz (8.1) Similarly,for equilibrium of forces in the y and z directions, Pyn =oyy cos ny +tyx cos nx tyz cos nz (8.2) Pan =Ou cos nz +tur cos nx tay cos ny (8.3) The resultant stress pn on the plane ABC is then given by pm=Vp层n+p品+p) (8.4) The normal stress on is given by resolution perpendicular to the face ABC, i.e. On Pxn cos nx Pyn cos ny Pin cos nz (8.5) and,by Pythagoras'theorem (Fig.8.4),the shear stress tn is given by tn =V(pi-2) (8.6) Ammmmro n C Fig.8.4.Normal,shear and resultant stresses on the plane ABC. It is often convenient and quicker to define the line of action of the resultant stress p by the direction cosines I'=cos(Pnx)=Pxn/Pn (8.7) m'=cos(Pny)=Pyn/pn (8.8) n'=cos(Pnz)=Pin/Pn (8.9)
58.3 Introduction to Advanced Elasticity Theory 223 the X direction due to a, is given by a,AS cos nx Stress components in the other axial directions will be similar in form. Thus, for equilibrium of forces in the X direction, h 3 pxnAS + F,AS- = cxxAScosnx + t,!,AScosny + t,,AScosnz As h + 0 (i.e. plane ABC passes through Q), the second term above becomes very small and can be neglected. The above equation then reduces to pxn = a, COS nx + txy COS ny + r,, COS nz (8.1) Similarly, for equilibrium of forces in the y and z directions, (8.2) i (8.3) pYn = uyy cos ny + tyx cos nx + tyz cos nz pm = a, cos nz + tu cos nx + tzy cos ny The resultant stress pn on the plane ABC is then given by Pn Jb:n + P;n + Pzn) (8.4) The normal stress a, is given by resolution perpendicular to the face ABC, i.e. and, by Pythagoras’ theorem (Fig. 8.4), the shear stress sn is given by = pxn COS nx + Pyn COS ny + Pzn COS nz rn = J(Pi-4) Fig. 8.4. Normal, shear and resultant stresses on the plane ABC. It is often convenient and quicker to define the line of action of the resu-.ant stress pn by the direction cosines 1’ = cos(pnx) = PxnIpn (8.7)
224 Mechanics of Materials 2 $8.4 The direction of the plane ABC being given by other direction cosines I cos nx,m cos ny,n cos ny It can be shown by simple geometry that 2+m2+n2=1and('2+(m)2+(n')2=1 Equations (8.1),(8.2)and (8.3)may now be written in two alternative ways. (a)Using the common symbol o for stress and relying on the double suffix notation to discriminate between shear and direct stresses: Pxn =Oxx cos nx Oxy cos ny +oxz cos nz (8.10) Pyn dyx cos nx oyy cos ny +oyz cos nz (8.11) Pan =xx cos nx oxy cos ny +oz cos nz (8.12) In each of the above equations the first suffix is common throughout,the second suffix on the right-hand-side terms are in the order x,y,z throughout,and in each case the cosine term relates to the second suffix.These points should aid memorisation of the equations. (b)Using the direction cosine form: Prm=oxxl+Oym+ox红n (8.13) Pyn dyxl +oyym oyzn (8.14) Pin Oul ogym +oun (8.15) Memory is again aided by the notes above,but in this case it is the direction cosines,l,m and n which relate to the appropriate second suffices x,y and z. Thus,provided that the direction cosines of a plane are known,together with the cartesian stress components at some point o on the plane,the direct,normal and shear stresses on the plane at o may be determined using,firstly,eqns.(8.13-15)and,subsequently, eqns.(8.4-6). Alternatively the procedure may be carried out graphically as will be shown in $8.9. 8.4.Direct,shear and resultant stresses on an oblique plane Consider again the oblique plane ABC having direction cosines /m and n,i.e.these are the cosines of the angle between the normal to plane and the x,y,z directions. In general,the resultant stress on the plane p will not be normal to the plane and it can therefore be resolved into two alternative sets of components. (a)In the co-ordinate directions giving components px P and pn,as shown in Fig.8.5, with values given by eqns.(8.13).(8.14)and (8.15). (b)Normal and tangential to the plane as shown in Fig.8.6.giving components,of on (normal or direct stress)and n(shear stress)with values given by eqns.(8.5)and (8.6). The value of the resultant stress can thus be obtained from either of the following equations: p后=品+品 (8.16)
224 Mechanics of Materials 2 88.4 The direction of the plane ABC being given by other direction cosines I = cosnx, tn = cosny, n = cosny It can be shown by simple geometry that l2 + m2 + n2 = 1 and (l’)’ + (m’)2 + (n’)* = 1 Equations (8.1), (8.2) and (8.3) may now be written in two alternative ways. (a) Using the cummon symbol cr for stress and relying on the double suffix notation to pxn = axx cos nx + uxy cos ny + axz cos nz (8. IO) pYn = uyx cos nx + ayy cos ny + uyz cos nz (8.1 1) pm = a, cos nx + azy cos ny + a, cos nz (8.12) In each of the above equations the first suffix is common throughout, the second suffix on the right-hand-side terms are in the order x, y, z throughout, and in each case the cosine term relates to the second suffix. These points should aid memorisation of the equations. discriminate between shear and direct stresses: (b) Using the direction cosine form: Pxn = uxxl + uxym + axzn pyn = uyxl+ uyym + uyzn pzn = uzrl + u,m + u,n (8.13) (8.14) (8. IS) Memory is again aided by the notes above, but in this case it is the direction cosines, 1, rn and n which relate to the appropriate second suffices x, y and z. Thus, provided that the direction cosines of a plane are known, together with the Cartesian stress components at some point Q on the plane, the direct, normal and shear stresses on the plane at Q may be determined using, firstly, eqns. (8.13-15) and, subsequently, eqns. (8.4-6). Alternatively the procedure may be carried out graphically as will be shown in $8.9. 8.4. Direct, shear and resultant stresses on an oblique plane Consider again the oblique plane ABC having direction cosines I, m and n, i.e. these are In general, the resultant stress on the plane pII will not be normal to the plane and it can the cosines of the angle between the normal to plane and the x, y. z directions. therefore be resolved into two alternative sets of Components. (a) In the co-ordinate directions giving components pI,, , pyn and p:,, , as shown in Fig. 8.5, with values given by eqns. (8.13), (8.14) and (8.15). (b) Normal and tangential to the plane as shown in Fig. 8.6, giving components, of a,, (normal or direct stress) and r,, (shear stress) with values given by eqns. (8.5) and (8.6). The value of the resultant stress can thus be obtained from either of the following equations: p,Z=d+d (8.16)
§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