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