CHAPTER 13 COMPLEX STRESSES Summary The normal stress o and shear stress t on oblique planes resulting from direct loading are a=a sin20 and t=sin 20 The stresses on oblique planes owing to a complex stress system are: normal stress =(ox+)+(ax-)cos 20+txy sin 20 shear stress =(x-)sin 20-tx cos 20 The principal stresses (i.e.the maximum and minimum direct stresses)are then o1=(ox+0,)+√[(ax-,2+4t3] 02=(ox+oy)-√[(ox-0,)2+4t] and these occur on planes at an angle e to the plane on which ax acts,given by either tan28=,2” (ox-0) or tan0=p-0x where p=a1,or o2,the planes being termed principal planes.The principal planes are always at 90 to each other,and the planes of maximum shear are then located at 45 to them. The maximum shear stress is tmx=2√/[(ox-0y)2+4r,]=(o1-02) In problems where the principal stress in the third dimension a3 either is known or can be assumed to be zero,the true maximum shear stress is then (greatest principal stress-least principal stress) Normal stress on plane of maximum shear =(+ Shear stress on plane of maximum direct stress (principal plane)=0 Most problems can be solved graphically by Mohr's stress circle.All questions which are capable of solution by this method have been solved both analytically and graphically. 13.1.Stresses on oblique planes Consider the general case,shown in Fig.13.1,of a bar under direct load F giving rise to stress o,vertically. 326
CHAPTER 13 COMPLEX STRESSES Summary The normal stress a and shear stress z on oblique planes resulting from direct loading are a = ay sin’ 8 and z = 30, sin 28 The stresses on oblique planes owing to a complex stress system are: normal stress = +(a, + ay) ++(ax - a,) cos 28 + zXy sin 28 shear stress = +(a, - cy) sin 28 - zXy cos 28 The principal stresses (i.e. the maximum and minimum direct stresses) are then a1 = *(a, + a,,) + ,J[ (a, -ay)’ + 45py] 0’ = *(a, + (ty) - $J[ (6, - o~)’ + 4~:,] and these occur on planes at an angle 0 to the plane on which a, acts, given by either where aP = alr or a’, the planes being termed principal planes. The principal planes are always at 90” to each other, and the planes of maximum shear are then located at 45” to them. The maximum shear stress is zmax= 3JC(~x-~,)2+4~~,l = a@, -a’) In problems where the principal stress in the third dimension u3 either is known or can be assumed to be zero, the true maximum shear stress is then +(greatest principal stress - least principal stress) Normal stress on plane of maximum shear = $(a, + a,) Shear stress on plane of maximum direct stress (principal plane) = 0 Most problems can be solved graphically by Mohr’s stress circle. All questions which are capable of solution by this method have been solved both analytically and graphically. 13.1. Stresses on oblique planes Consider the general case, shown in Fig. 13.1, of a bar under direct load F giving rise to stress by vertically. 326
§13.2 Complex Stresses 327 -Unit depth Fig.13.1.Bar subjected to direct stress,showing stresses acting on any inclined plane. Let the block be of unit depth;then considering the equilibrium of forces on the triangular portion ABC: resolving forces perpendicular to BC, og×BC×1=g,×AB×1×sin6 But AB BC sin 0, ae=a,sin20 (13.1) Now resolving forces parallel to BC, tg×BC×1=o,×AB×1Xcos6 Again AB BC sin 0. Te =y sin 0 cos0 =t,sin 20 (13.2) The stresses on the inclined plane,therefore,are not simply the resolutions of oy perpendicular and tangential to that plane.The direct stress has a maximum value of when6=90°whilst the shear stress te has a maximum value of to,whenθ=45°， Thus any material whose yield stress in shear is less than half that in tension or compression will yield initially in shear under the action of direct tensile or compressive forces. This is evidenced by the typical"cup and cone"type failure in tension tests of ductile specimens such as low carbon steel where failure occurs initially on planes at 45 to the specimen axis.Similar effects occur in compression tests on,for example,timber where failure is again due to the development of critical shear stresses on 45 planes. 13.2.Material subjected to pure shear Consider the element shown in Fig.13.2 to which shear stresses have been applied to the sides AB and DC.Complementary shear stresses of equal value but of opposite effect are then set up on sides AD and BC in order to prevent rotation of the element.Since the applied and complementary shears are of equal value on the x and y planes,they are both given the symbol tx
$13.2 c Complex Stresses 327 t Fig. 13.1. Bar subjected to direct stress, showing stresses acting on any inclined plane. Let the block be of unit depth; then considering the equilibrium of forces on the triangular resolving forces perpendicular to BC, portion ABC: oo x BC x 1 = Q,, x AB x 1 x sin8 But AB = BC sin 8, .. bg = uy sin2 8 (13.1) Now resolving forces parallel to BC, T,x~cxi =o,,x~~xixcose Again AB = BC sin 8, .. to = sin 8 cos 8 = 30, sin 28 The stresses on the inclined plane, therefore, are not simply the resolutions of o, perpendicular and tangential to that plane. The direct stress uo has a maximum value of Q, when 8 = 90" whilst the shear stress q, has a maximum value of 30, when 8 = 45". Thus any material whose yield stress in shear is less than half that in tension or compression will yield initially in shear under the action of direct tensile or compressive forces. This is evidenced by the typical "cup and cone" type failure in tension tests of ductile specimens such as low carbon steel where failure occurs initially on planes at 45" to the specimen axis. Similar effects occur in compression tests on, for example, timber where failure is again due to the development of critical shear stresses on 45" planes. (1 3.2) 13.2. Material subjected to pure shear Consider the element shown in Fig. 13.2 to which shear stresses have been applied to the sides AB and DC. Complementary shear stresses of equal value but of opposite effect are then set up on sides AD and BC in order to prevent rotation of the element. Since the applied and complementary shears are of equal value on the x and y planes, they are both given the symbol rXy
328 Mechanics of Materials S13.2 A y Fig.13.2.Stresses on an element subjected to pure shear. Consider now the equilibrium of portion PBC. Resolving normal to PC assuming unit depth, dox PC=txy×BC sin0+txy×PB cos0 =txy×PC cos sinθ+txy×PC sin0 cos0 o=txy sin 20 (13.3) The maximum value of ce is txy when 6=45. Similarly,resolving forces parallel to PC, tox PC txy x PB sin 0-txy BC cos0 =txw×PC sin20-ty×PCcos20 to=-txy Cos 20 (13.4) The negative sign means that the sense of t is opposite to that assumed in Fig.13.2. The maximum value of te is xy when=0or 90and it has a value of zero when=45, i.e.on the planes of maximum direct stress. Further consideration ofegns.(13.3)and(13.4)shows that the system of pure shear stresses produces an equivalent direct stress system as shown in Fig.13.3,one set compressive and one tensile,each at 45 to the original shear directions,and equal in magnitude to the applied shear. G-Twy 451 0=--Txy Fig.13.3.Direct stresses due to shear. This has great significance in the measurement of shear stresses or torques on shafts using strain gauges where the gauges are arranged to record the direct strains at 45 to the shaft axis. Practical evidence of the theory is also provided by the failure of brittle materials in shear. A shaft of a brittle material subjected to torsion will fail under direct stress on planes at 45to the shaft axis.(This can be demonstrated easily by twisting a piece of blackboard chalk in
328 Mechanics of Materials $13.2 Fig. 13.2. Stresses on an element subjected to pure shear. Consider now the equilibrium of portion PBC. Resolving normal to PC assuming unit depth, no x PC = zxy x BC sin8+zxy x PB cos8 (13.3) = zxy x PC cos 8 sin 8 + zxy x PC sin 8 cos 8 .. The maximum value of no is zxy when 8 = 45". Similarly, resolving forces parallel to PC, q, = zXy sin 28 z, x PC = zxy x PB sin 8 - zxy BC cos 8 = zxy x PC sin2 8 - zXy x PC cos2 e .. -zxycos28 (13.4) The negative sign means that the sense of To is opposite to that assumed in Fig. 13.2. The maximum value of 7, is zxy when 8 = 0" or 90" and it has a value of zero when 8 = 45", i.e. on the planes of maximum direct stress. Further consideration of eqns. (13.3) and (13.4) shows that the system of pure shear stresses produces an equivalent direct stress system as shown in Fig. 13.3, one set compressive and one tensile, each at 45" to the original shear directions, and equal in magnitude to the applied shear. XY Fig. 13.3. Direct stresses due to shear. This has great sign@ance in the measurement of shear stresses or torques on shafts using strain gauges where the gauges are arranged to record the direct strains at 45" to the shaft axis. Practical evidence of the theory is also provided by the failure of brittle materials in shear. A shaft of a brittle material subjected to torsion will fail under direct stress on planes at 45" to the shaft axis. (This can be demonstrated easily by twisting a piece of blackboard chalk in
§13.3 Complex Stresses 329 one's hands;see Fig.8.8a on page 185.)Tearing of a wet cloth when it is being wrung out is also attributed to the direct stresses introduced by the applied torsion. 13.3.Material subjected to two mutually perpendicular direct stresses Consider the rectangular element of unit depth shown in Fig.13.4 subjected to a system of two direct stresses,both tensile,at right angles,ox and o,. For equilibrium of the portion ABC,resolving perpendicular to AC, og×AC×1=ox×BC×1×cos0+o,×AB×1×sin0 =ox×AC cos29+o,×AC sin20 0g=ox(1+cos20+o,(1-cos20) i.e. 00=(ax+0,)+(ox-c,)c0s20 (13.5) 6 Unit depth Fig.13.4.Element from a material subjected to two mutually perpendicular direct stresses. Resolving parallel to AC: ta×AC×1=ox×BC×1×sin8-o,×AB×1Xcos0 te=x cos sin0-ay cos 0 sin0 tg=(ox-o,）sin20 (13.6) The maximum direct stress will equal ox or whichever is the greater,when=0or 90. The maximum shear stress in the plane of the applied stresses (see $13.8)occurs when 0=45°， i.e. tmx=是(ox-0,) (13.7) 13.4.Material subjected to combined direct and shear stresses Consider the complex stress system shown in Fig.13.5 acting on an element of material. The stresses and may be compressive or tensile and may be the result of direct forces or bending.The shear stresses may be as shown or completely reversed and occur as a result of either shear forces or torsion
$13.3 Complex Stresses 329 one's hands; see Fig. 8.8a on page 185.) Tearing of a wet cloth when it is being wrung out is also attributed to the direct stresses introduced by the applied torsion. 13.3. Material subjected to two mutually perpendicular direct stresses Consider the rectangular element of unit depth shown in Fig. 13.4 subjected to a system of For equilibrium of the portion ABC, resolving perpendicular to AC, two direct stresses, both tensile, at right angles, ax and ay. box ACx 1 = a, x BC x 1 xcos8+oY x ABx 1 x sin8 = a, x AC cos' 8 +ay x AC sin'8 .. = +ox (1 + COS 28) + icy (1 - cos 28) i.e. ug = ~(ux+uy)+~(ux-ubg)c~~2e (1 3.5) t i Fig. 13.4. Element from a material subjected to two mutually perpendicular direct stresses. Resolving parallel to AC: zo x AC x 1 = a, x BC x 1 x sine-a, x AB x 1 x cos8 .. zo = a, cos 8 sin 8 - ay cos 8 sin 8 re = $(ux - uy) sin 28 (13.6) The maximum direct stress will equal a, or by, whichever is the greater, when 8 = 0 or 90". The maximum shear stress in the plane of the applied stresses (see $13.8) occurs when e = 450, i.e. rm.x = 4 (ax - by) (13.7) 13.4. Material subjected to combined direct and shear stresses Consider the complex stress system shown in Fig. 13.5 acting on an element of material. The stresses a, and ay may be compressive or tensile and may be the result of direct forces or bending. The shear stresses may be as shown or completely reversed and occur as a result of either shear forces or torsion
330 Mechanics of Materials §13.4 y Fig.13.5.Two-dimensional complex stress system. The diagram thus represents a complete stress system for any condition of applied load in two dimensions and represents an addition of the stress systems previously considered in §§13.2and13.3. The formulae obtained in these sections may therefore be combined to give o=(ax+a)+(ax-a)cos 20+ixy sin 20 (13.8) and to=(ox-a)sin 20-txy cos 20 (13.9) The maximum and minimum stresses which occur on any plane in the material can now be determined as follows: For oe to be a maximum or minimum doe=0 de Now o=(ox+)+(ax-G)cos 20+txy sin 20 do =-(-,sin 20+2tx cos20-0 do or tan 20 Lisy (13.10) (ax-0,) from Fig.13.6 sin 20 = 2txy √/[(ox-o,)2+4t,] (x-0) cos 20=- √/[ax-,)2+4t] (G,-0+4Ty 2y 28 (-cy} Fig.13.6
330 Mechanics of Materials $1 3.4 t Fig. 13.5. Two-dimensional complex stress system The diagram thus represents a complete stress system for any condition of applied load in two dimensions and represents an addition of the stress systems previously considered in Gg13.2 and 13.3. The formulae obtained in these sections may therefore be combined to give ug = +(u, + u,) ++(u, - u,) cos 28 +7,, sin 28 (1 3.8) and zg= ~(u,-u,)sin28--z,,cos28 (1 3.9) determined as follows: The maximum and minimum stresses which occur on any plane in the material can now be For ag to be a maximum or minimum Now ~ =o 40 dg = +(a, + a,) ++(a, - a,) COS 20 + T,~ sin 20 3 = - (a, - ay) sin 20 + 25,, cos 20 = o de .. or .'. from Fig. 13.6 sin20 = - JW, - + 4e,1 (1 3.10) (u,-u,) Fig. 13.6
s13.5 Complex Stresses 331 Therefore substituting in eqn.(13.8),the maximum and minimum direct stresses are given by 10r2=ox+o,)+生亿：-，a:-0,) Txy×2ry √/[(o.-o,)2+4t]'√[(ox-o,)2+4t] =(ox+o,)±√[(ox-o,)2+4t,] (13.11) These are then termed the principal stresses of the system. The solution of eqn.(13.10)yields two values of 20 separated by 180,i.e.two values of separated by 90.Thus the two principal stresses occur on mutually perpendicular planes termed principal planes,and substitution for e from eqn.(13.10)into the shear stress expression eqn.(13.9)will show that to=0 on the principal planes. The complex stress system of Fig.13.5 can now be reduced to the equivalent system of principal stresses shown in Fig.13.7. Principal planes Fig.13.7.Principal planes and stresses. From eqn.(13.7)the maximum shear stress present in the system is given by t=2(o1-02) (13.12) =t√[(ox-o,)2+4,] (13.13) and this occurs on planes at 45 to the principal planes. This result could have been obtained using a similar procedure to that used for determining the principal stresses,i.e.by differentiating expression (13.9),equating to zero and substituting the resulting expression for 6. 13.5.Principal plane inclination in terms of the associated principal stress It has been stated in the previous section that expression (13.10),namely tan 20 2tx (ox-0) yields two values of 0,i.e.the inclination of the two principal planes on which the principal stresses a;and g2 act.It is uncertain,however,which stress acts on which plane unless eqn. (13.8)is used,substituting one value of 6 obtained from eqn.(13.10)and observing which one of the two principal stresses is obtained.The following alternative solution is therefore to be preferred
$1 3.5 Complex Stresses 33 1 Therefore substituting in eqn. (13.Q the maximum and minimum direct stresses are given by These are then termed the principal stresses of the system. The solution of eqn. (13.10) yields two values of 28 separated by 180", i.e. two values of 8 separated by 90". Thus the two principal stresses occur on mutually perpendicular planes termed principal planes, and substitution for 8 from eqn. (13.10) into the shear stress expression eqn. (13.9) will show that z, = 0 on the principal planes. The complex stress system of Fig. 13.5 can now be reduced to the equivalent system of principal stresses shown in Fig. 13.7. Fig. 13.7. Principal planes and stresses. From eqn. (13.7) the maximum shear stress present in the system is given by z,= +@l-oZ) (1 3.12) = &JC(n, - @,I2 + 421y1 ( 1 3.1 3) and this occurs on planes at 45" to the principal planes. This result could have been obtained using a similar procedure to that used for determining the principal stresses, i.e. by differentiating expression (13.9), equating to zero and substituting the resulting expression for 8. 13.5. Principal plane inclination in terms of the associated principal stress It has been stated in the previous section that expression (13.10), namely 25xy (ax - a,) tan 28 = ~ yields two values of 8, i.e. the inclination of the two principal planes on which the principal stresses o1 and a2 act. It is uncertain, however, which stress acts on which plane unless eqn. (13.8) is used, substituting one value of 8 obtained from eqn. (13.10) and observing which one of the two principal stresses is obtained. The following alternative solution is therefore to be preferred
332 Mechanics of Materials §13.6 Consider once again the equilibrium of a triangular block of material of unit depth (Fig.13.8);this time AC is a principal plane on which a principal stress p acts,and the shear stress is zero (from the property of principal planes). ÷0 Fig.13.8. Resolving forces horizontally, (ox×BC×1)+(txy×AB×1)=(op×AC×1)cos8 Ox+txy tan0=op tan0=p-ox (13.14) 下y Thus we have an equation for the inclination of the principal planes in terms of the principal stress.If,therefore,the principal stresses are determined and substituted in the above equation,each will give the corresponding angle of the plane on which it acts and there can then be no confusion. The above formula has been derived with two tensile direct stresses and a shear stress system,as shown in the figure;should any of these be reversed in action,then the appropriate minus sign must be inserted in the equation. 13.6.Graphical solution-Mohr's stress circle Consider the complex stress system of Fig.13.5 (p.330).As stated previously this represents a complete stress system for any condition of applied load in two dimensions. In order to find graphically the direct stress and shear stress te on any plane inclined at 0 to the plane on which a acts,proceed as follows: (1)Label the block ABCD. (2)Set up axes for direct stress (as abscissa)and shear stress (as ordinate)(Fig.13.9). (3)Plot the stresses acting on two adjacent faces,e.g.AB and BC,using the following sign conventions: direct stresses:tensile,positive;compressive,negative; shear stresses:tending to turn block clockwise,positive;tending to turn block counterclockwise,negative. This gives two points on the graph which may then be labelled AB and BC respectively to denote stresses on these planes
332 Mechanics of Materials 513.6 Consider once again the equilibrium of a triangular block of material of unit depth (Fig. 13.8); this time AC is a principal plane on which a principal stress up acts, and the shear stress is zero (from the property of principal planes). UY t Fig. 13.8. Resolving forces horizontally, (1 3.14) Thus we have an equation for the inclination of the principal planes in terms of the principal stress. If, therefore, the principal stresses are determined and substituted in the above equation, each will give the corresponding angle of the plane on which it acts and there can then be no confusion. The above formula has been derived with two tensile direct stresses and a shear stress system, as shown in the figure; should any of these be reversed in action, then the appropriate minus sign must be inserted in the equation. 13.6. Graphical solution - Mohr’s stress circle Consider the complex stress system of Fig. 13.5 (p. 330). As stated previously this represents a complete stress system for any condition of applied load in two dimensions. In order to find graphically the direct stress and shear stress tg on any plane inclined at 8 to the plane on which CT, acts, proceed as follows: (1) Label the block ABCD. (2) Set up axes for direct stress (as abscissa) and shear stress (as ordinate) (Fig. 13.9). (3) Plot the stresses acting on two adjacent faces, e.g. AB and BC, using the following sign conventions: direct stresses: tensile, positive; compressive, negative; shear stresses: tending to turn block clockwise, positive; tending to turn block counterclockwise, negative. This gives two points on the graph which may then be labelled AB and E respectively to denote stresses on these planes
§13.6 Complex Stresses 333 AB 28 No M B=28 02 BC Fig.13.9.Mohr's stress circle. (4)Join AB and BC. (5)The point P where this line cuts the a axis is then the centre of Mohr's circle,and the line is the diameter;therefore the circle can now be drawn. Every point on the circumference of the circle then represents a state of stress on some plane through C. Proof Consider any point on the circumference of the circle,such that PO makes an angle 20 with BC,and drop a perpendicular from O to meet the o axis at N. Coordinates of O: ON OP+PN=(ax+a)+R cos(20-B) =(x+)+R cos 20 cos B+R sin 20 sin B But R cosB=(ax-)and R sinB=txy ON =(x+a)+(ax-)cos 20+txy sin 20 On inspection this is seen to be eqn.(13.8)for the direct stress on the plane inclined at to BC in Fig.13.5. Similarly, QN R sin (20-B) =R sin 20 cos B-R cos 20 sin B =(ox-a)sin 20-txy cos 20 Again,on inspection this is seen to be egn.(13.9)for the shear stress te on the plane inclined at 0 to BC
$1 3.6 Complex Stresses 333 I I c v 01 1 I I Fig. 13.9. Mohr’s stress circle. Join AB and z. The point P where this line cuts the a axis is then the centre of Mohr’s line is the diameter; therefore the circle can now be drawn. circle, and the Every point on the circumference of the circle then represents a state of stress on some plane through C. Proof Consider any point Q on the circumference of the circle, such that PQ makes an angle 28 with E, and drop a perpendicular from Q to meet the a axis at N. Coordinates of Q: ON = OP+PN = ~(0,+a,)+R~0~(28-fl) = $(ax + a,) + R cos 28 cos p + R sin 28 sin p But R cos p = +(a, - a,) and R sin p = T,~ .. On inspection this is seen to be eqn. (13.8) for the direct stress BC in Fig. 13.5. ON = +(ax + a,) ++(a, - a,)cos 28 + T,, sin 28 on the plane inclined at 8 to Similarly, QN = R sin (28 - p) = Rsin28cos~-Rcos28sin~ = +(ax - a,) sin 28 - T,~ cos 28 Again, on inspection this is seen to be eqn. (13.9) for the shear stress TO on the plane inclined at t) to BC
334 Mechanics of Materials §13.7 Thus the coordinates of Q are the normal and shear stresses on a plane inclined at 0 to BC in the original stress system. N.B.-Single angle BCPO is 20 on Mohr's circle and not 0,it is evident that angles are doubled on Mohr's circle.This is the only difference,however,as they are measured in the same direction and from the same plane in both figures (in this case counterclockwise from BC). Further points to note are: (1)The direct stress is a maximum when O is at M,i.e.OM is the length representing the maximum principal stresso;and 20,gives the angle of the plane from BC.Similarly, OL is the other principal stress. (2)The maximum shear stress is given by the highest point on the circle and is represented by the radius of the circle.This follows since shear stresses and complementary shear stresses have the same value;therefore the centre of the circle will always lie on the a axis midway between a.and a,. (3)From the above point the direct stress on the plane of maximum shear must be midway between a,and a,,i.e.(x+,). (4)The shear stress on the principal planes is zero. (5)Since the resultant of two stresses at 90can be found from the parallelogram of vectors as the diagonal,as shown in Fig.13.10,the resultant stress on the plane at 0 to BC is given by 00 on Mohr's circle. Fig.13.10.Resultant stress (a,)on any plane. The graphical method of solution of complex stress problems using Mohr's circle is a very powerful technique since all the information relating to any plane within the stressed element is contained in the single construction.It thus provides a convenient and rapid means of solution which is less prone to arithmetical errors and is highly recommended. With the growing availability and power of programmable calculators and microcom- puters it may be that the practical use of Mohr's circle for the analytical determination of stress (and strain-see Chapter 14)values will become limited.It will remain,however,a highly effective medium for the teaching and understanding of complex stress systems. A free-hand sketch of the Mohr circle construction,for example,provides a convenient mechanism for the derivation (by simple geometric relationships)of the principal stress equations(13.11)or of the equations for the shear and normal stresses on any inclined plane in terms of the principal stresses as shown in Fig.13.11. 13.7.Alternative representations of stress distributions at a point The way in which the stress at a point varies with the angle at which a plane is taken through the point may be better understood with the aid of the following alternative graphical representations
334 Mechanics of Materials 613.7 Thus the coordinates of Q are the normal and shear stresses on a plane inclined at 8 to BC in the original stress system. N.B.-Single angle ZPQ is 28 on Mohr’s circle and not 8, it is evident that angles are doubled on Mohr’s circle. This is the only difference, however, as they are measured in the same direction and from the same plane in both figures (in this case counterclockwise from BC ). - Further points to note are: (1) The direct stress is a maximum when Q is at M, i.e. OM is the length representing the maximum principal stress a1 and 28, gives the angle of the plane 8, from BC. Similarly, OL is the other principal stress. (2) The maximum shear stress is given by the highest point on the circle and is represented by the radius of the circle. This follows since shear stresses and complementary shear stresses have the same value; therefore the centre of the circle will always lie on the a axis midway between a, and a,. (3) From the above point the direct stress on the plane of maximum shear must be midway between a, and a,,, i.e. $(a, + a,). (4) The shear stress on the principal planes is zero. (5) Since the resultant of two stresses at 90” can be found from the parallelogram of vectors as the diagonal, as shown in Fig. 13.10, the resultant stress on the plane at 8 to BC is given by OQ on Mohr’s circle. Fig. 13.10. Resultant stress (8,) on any plane. The graphical method of solution of complex stress problems using Mohr’s circle is a very powerful technique since all the information relating to any plane within the stressed element is contained in the single construction. It thus provides a convenient and rapid means of solution which is less prone to arithmetical errors and is highly recommended. With the growing availability and power of programmable calculators and microcomputers it may be that the practical use of Mohr’s circle for the analytical determination of stress (and strain-see Chapter 14) values will become limited. It will remain, however, a highly effective medium for the teaching and understanding of complex stress systems. A free-hand sketch of the Mohr circle construction, for example, provides a convenient mechanism for the derivation (by simple geometric relationships) of the principal stress equations (13.1 1) or of the equations for the shear and normal stresses on any inclined plane in terms of the principal stresses as shown in Fig. 13.11. 13.7. Alternative representations of stress distributions at a point The way in which the stress at a point vanes with the angle at which a plane is taken through the point may be better understood with the aid of the following alternative graphical representations
§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 twodimensional 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
