Engineering Mechanics of Solids by Popov Chapter 8: Transformation of Stress and Strain; Yield and Fracture Criteria

1. Chapter 8
Transformation of Stress and Strain;
Yield and Fracture Criteria
Mechanics of Solids

2. Fig. 1: State of stress at a point on different planes
Fig. 2: Representations of stresses acting on an element

3. Part A- Transformation of Stress
• Transformation of stresses in 2 dimensional problem can be computed
using Fig.3,
Fig. 3: Derivation of stress transformation on an inclined plane

4. 𝐹𝑥′ = 0 𝜎𝑥′ 𝑑𝐴 = 𝜎𝑥 𝑑𝐴 cos 𝜃 cos 𝜃 + 𝜎 𝑦 𝑑𝐴 sin 𝜃 sin 𝜃
+𝜏 𝑥𝑦 𝑑𝐴 cos 𝜃 sin 𝜃 + 𝜏 𝑥𝑦 𝑑𝐴 sin 𝜃 cos 𝜃
𝜎𝑥′ = 𝜎𝑥cos2 𝜃 + 𝜎 𝑦sin2 𝜃 + 2𝜏 𝑥𝑦 sin 𝜃 cos 𝜃
= 𝜎𝑥
1+cos 2𝜃
2
+ 𝜎 𝑦
1−cos 2𝜃
2
+ 𝜏 𝑥𝑦 sin 2𝜃
𝜎𝑥
′ =
𝜎 𝑥+𝜎 𝑦
2
+
𝜎 𝑥−𝜎 𝑦
2
cos 2𝜃 + 𝜏 𝑥𝑦 sin 2𝜃
• Similarly from 𝐹𝑦′ = 0,
𝜏 𝑥′𝑦′ = −
𝜎 𝑥−𝜎 𝑦
2
sin 2𝜃 + 𝜏 𝑥𝑦 cos 2𝜃
• Replacing 𝜃 by 𝜃 + 90° gives the normal stress in the direction of the 𝑦′
axis.
𝜎 𝑦
′
=
𝜎 𝑥+𝜎 𝑦
2
−
𝜎 𝑥−𝜎 𝑦
2
cos 2𝜃 − 𝜏 𝑥𝑦 sin 2𝜃

5. 𝜎𝑥′ + 𝜎 𝑦′ = 𝜎𝑥 + 𝜎 𝑦
• In plane strain problems, where 𝜀 𝑧 = 𝛾𝑧𝑥 = 𝛾𝑧𝑦 = 0, a normal stress
𝜎𝑧 can also develop.
𝜎𝑧 = 𝜈 𝜎𝑥 + 𝜎 𝑦
Principal Stresses in Two-Dimensional Problems
• To find the plane for a maximum or a minimum normal stresses,
𝑑𝜎 𝑥′
𝑑𝜃
= −
𝜎 𝑥−𝜎 𝑦
2
2 sin 2𝜃 + 2𝜏 𝑥𝑦 cos 2𝜃 = 0
tan 2𝜃1 =
2𝜏 𝑥𝑦
𝜎 𝑥−𝜎 𝑦

6. • This equation has two roots, giving maximum and minimum normal
stresses (principal stresses).
Fig. 4: Angle functions for principal stresses
• Using the equation of angle for principal planes,

7. 𝜎𝑥′ 𝑚𝑎𝑥
𝑚𝑖𝑛
= 𝜎1 𝑜𝑟 2 =
𝜎 𝑥+𝜎 𝑦
2
±
𝜎 𝑥−𝜎 𝑦
2
2
+ 𝜏 𝑥𝑦
2
• Where the positive sign in front of the square root must be used to
obtain 𝜎1 and the negative sign to obtain 𝜎2.
Maximum Shear Stresses in Two-Dimensional Problems
• We know that,
𝜏 𝑥′𝑦′ = −
𝜎 𝑥−𝜎 𝑦
2
sin 2𝜃 + 𝜏 𝑥𝑦 cos 2𝜃
• Differentiating this equation with respect to 𝜃 and setting the
derivative equal to zero to find maximum or minimum shear stresses.

8. tan 2𝜃2 = −
𝜎 𝑥−𝜎 𝑦 2
𝜏 𝑥𝑦
• The above equation has two roots. The two planes defined by this
equation are mutually perpendicular.
• The value of tan 2𝜃2 is negative reciprocal of tan 2𝜃1. The angles that
locates the planes of maximum or minimum shear stress form angles
of 45° with the planes of the principal stresses.
• A substitution of the sine and cosine functions corresponding to
double angle given by above equation and determined in a manner
analogous to that in Fig. 4 gives the maximum and the minimum
values of the shear stresses.
𝜏 𝑚𝑎𝑥
𝑚𝑖𝑛
= ±
𝜎 𝑥−𝜎 𝑦
2
2
+ 𝜏 𝑥𝑦
2

9. • A positive shear stress indicates that it acts in the direction assumed in
Fig. 1(b), and vice versa.
• The maximum shear stresses act on planes that are usually not free of
normal stresses.
• The normal stresses that act on the planes of the maximum shear
stresses are
𝜎′ =
𝜎 𝑥+𝜎 𝑦
2
• If 𝜎𝑥 and 𝜎 𝑦 are the principal stress, 𝜏 𝑥𝑦 is zero.
𝜏 𝑚𝑎𝑥 =
𝜎1−𝜎2
2
• The results of this analysis are displayed in Fig. 5. the shear stresses act
toward the diagonal 𝐷𝐹 in the direction of the principal tensile stresses,
Fig. 5(a).

10. Mohr’s Circle of Stress for 2-D Problems
• Transformed stresses given as 𝜎𝑥′ and 𝜏 𝑥′𝑦′ represent a circle written in
parametric form.
𝜎𝑥
′ −
𝜎 𝑥+𝜎 𝑦
2
=
𝜎 𝑥−𝜎 𝑦
2
cos 2𝜃 + 𝜏 𝑥𝑦 sin 2𝜃
𝜏 𝑥′𝑦′ = −
𝜎 𝑥−𝜎 𝑦
2
sin 2𝜃 + 𝜏 𝑥𝑦 cos 2𝜃
• By squaring both these equations, adding, and simplifying,
𝜎𝑥
′
−
𝜎 𝑥+𝜎 𝑦
2
2
+ 𝜏 𝑥′𝑦′
2
=
𝜎 𝑥−𝜎 𝑦
2
2
+ 𝜏 𝑥𝑦
2
• The above equation represents a circle of radius
𝜎 𝑥−𝜎 𝑦
2
2
+ 𝜏 𝑥𝑦
2
with its center at +
𝜎 𝑥+𝜎 𝑦
2
, 0 .

11. • The ordinate of a point on the circle is the shear stress 𝜏 𝑥′𝑦′ and the
abscissa is the normal stress 𝜎𝑥′.
• The circle so constructed is called a circle of stress or Mohr’s circle of
stress.
Fig. 5

12. • The coordinates for point 𝐴 on the circle correspond to the stresses in
Fig. 5(a) on the right face of the element. For this face of the element,
θ = 0°, i.e., the 𝑥𝑦 and the 𝑥′𝑦′ axes coincide, and 𝜎𝑥′ = 𝜎𝑥 and
𝜏 𝑥′𝑦′ = 𝜏 𝑥𝑦.
• The positive directions for these stresses coincide with the positive
directions of the axes.
• Since 𝐴𝐷 𝐶𝐷 = 𝜏 𝑥𝑦 𝜎𝑥 − 𝜎 𝑦 2 , the angle 𝐴𝐶𝐷 is equal to 2𝜃1.
• The coordinates for the conjugate point 𝐵 correspond to the stresses
in Fig. 5(a) on the upper surface of the element.
• The same reasoning applies to any other orientation of an element,
Fig. 5(b). A pair of conjugate points 𝐽 and 𝐾 can always be found on
the circle to give the corresponding stresses, Fig. 8(c).

13. • Important observations regarding the state of stress at a point:
1. The largest possible normal stress is 𝜎1; the smallest is 𝜎2. No shear
stresses exist together with either one of these principal stresses.
2. The largest shear stress 𝜏 𝑚𝑎𝑥 is numerically equal to the radius of the
circle, 𝜎1 − 𝜎2 2. A normal stress equal to 𝜎1 + 𝜎2 2 acts on
each of the planes of maximum shear stress.
3. If 𝜎1 = 𝜎2, Mohr’s circle degenerates into a point, and no shear
stresses at all develop in the 𝑥𝑦 plane.
4. If 𝜎𝑥 + 𝜎 𝑦 = 0, the center of Mohr’s circle coincides with the origin
of the 𝜎𝜏 coordinates, and the state of pure shear exists.
5. The sum of normal stresses on any two mutually perpendicular
planes is invariant, i.e.,
𝜎𝑥 + 𝜎 𝑦 = 𝜎1 + 𝜎2 = 𝜎𝑥′ + 𝜎 𝑦′ = constant

14. Construction of Mohr’s Circle for Stress
Transformation
Fig. 6: Construction of Mohr’s circle of stress.
stresses on arbitrary and principal Normal
and shear planes are shown in (c) and (d),
respectively.

15. • The state of stress in the 𝑥𝑦 coordinate system is shown in Fig. 6(a).
The objective is to transform the given stresses to those in the rotated
set of 𝑥′𝑦′ axes as shown in Figs. 6(a) and (b) by using Mohr’s circle.
• If 𝜏 𝑥𝑦 > 0, it is plotted downwards at 𝜎𝑥, and
if 𝜏 𝑥𝑦 < 0, it is plotted upwards at 𝜎𝑥.
• This in effect amounts to directing the positive τ axis downward, as
shown in Fig. 6(c).
• The coordinates of 𝜎𝑥 and 𝜏 𝑥𝑦 locate the governing point 𝐴 𝐶 on the
circle. This point corresponds to point 𝐴 in the earlier construction, see
Fig. 5.
• Point 𝐵 𝐶, conjugate to point 𝐴 𝐶, can be located on the circle as shown
in Fig. 6(c). The double angle 2θ follows from geometry.

16. • The diameter 𝐴 𝐶 𝐵 𝐶 is rotated through an angle 2θ in the same sense
that the 𝑥′ axis is rotated through the angle θ with respect to the 𝑥
axis.
• The new point 𝐴 𝐶′ determines the stresses 𝜎𝑥′ and 𝜏 𝑥′𝑦′ on the right
hand face of the element in Fig. 6(b).
• For the case shown, the shear stress 𝜏 𝑥′𝑦′ is negative, since at 𝜎𝑥′ it is
above the σ axis. Similar considerations apply to the conjugate point
𝐵 𝐶′ defining the stresses on the plane normal to the 𝑦′ axis.
• The expressions for 𝜎𝑥′ and 𝜏 𝑥′𝑦′ can be formulated from the
construction of the Mohr’s circle shown in Fig. 6(c).
• Procedure for determining the principal normal stress is shown in Fig.
6(d). The principal shear stresses are given by the coordinates of the
point on a circle at their extreme values on the 𝜏 axis.

17. Part B- Transformation of Strain
Strains in Two Dimensions
Fig. 7: Strains are determined from relative deformations

18. • In studying the strains at a point, only the relative displacement of the
adjoining points is of importance. Translation and rotation of an
element as a whole are of no consequence since these displacements
are rigid body displacements.
• If the extensional strain of a diagonal 𝑑𝑠 of the original element in Fig.
7(a) is being studied, the element in its deformed condition can be
brought back for comparison purposes, Fig. 7(c).
• For the small strains and the rotations considered, elongation 𝑑∆ in the
direction of the diagonal, is essentially the same.

19. Transformation of Strains in 2-D
• The normal strains 𝜀 𝑥 and 𝜀 𝑦 corresponding to elongations in the 𝑥
and 𝑦 directions, respectively, are taken positive. The element
distorted by positive shear strain will be taken as that shown in Fig.
8(a).
• The new 𝑥′ 𝑦′ system of axes is related to the 𝑥𝑦 axes as in Fig. 8(b).
• The displacement of point 𝐴 in the 𝑥 direction is 𝐴𝐴′
= 𝜀 𝑥 𝑑𝑥; in the 𝑦
direction, 𝐴′𝐴′′ = 𝜀 𝑦 𝑑𝑦.
• Assuming that the shear strain causes the horizontal displacement
shown in Fig. 8(a), 𝐴′′𝐴′′′
= 𝛾𝑥𝑦 𝑑𝑦.
• By projecting these displacements onto the 𝑥′ axis, displacement of
point 𝐴 along the 𝑥′ axis is found.

20. Fig. 8: Exaggerated deformations of elements for deriving strains along new axes

21. 𝜀 𝑥′ 𝑑𝑥′
= 𝐴𝐴′
cos θ + 𝐴′𝐴′′
sin θ + 𝐴′′𝐴′′′
cos θ
𝜀 𝑥′ 𝑑𝑥′ = 𝜀 𝑥
𝑑𝑥
𝑑𝑥′
cos θ + 𝜀 𝑦
𝑑𝑦
𝑑𝑥′
cos θ + 𝛾𝑥𝑦
𝑑𝑦
𝑑𝑥′
cos θ
• Since, however, 𝑑𝑥 𝑑𝑥′ = cos θ and 𝑑𝑦 𝑑𝑥′ = sin θ,
𝜀 𝑥′ = 𝜀 𝑥 cos2
θ + 𝜀 𝑦 sin2
θ + 𝛾𝑥𝑦 sin θ cos θ
𝜀 𝑥′ =
𝜀 𝑥+𝜀 𝑦
2
+
𝜀 𝑥−𝜀 𝑦
2
cos 2θ +
𝛾 𝑥𝑦
2
sin 2θ
• Above expression is for normal strain transformation in a plane in an
arbitrary direction defined by the 𝑥′ axis.
• Consider an element 𝑂𝐴𝐶𝐵 with sides 𝑂𝐴 and 𝑂𝐵 directed along the
𝑥′ and 𝑦′ axes, as shown in Fig. 8(b).

22. • The shear strain for this element is the change in angle 𝐴𝑂𝐵, 𝛼 + 𝛽.
• For small deformations, the small angle 𝛼 can be determined by
projecting the displacements 𝐴𝐴′, 𝐴′𝐴′′, and 𝐴′′𝐴′′′ onto a normal to
𝑂𝐴 and dividing this quantity by 𝑑𝑥′.
𝛼 ≈ tan 𝛼 =
−𝐴𝐴′ sin θ+𝐴𝐴′ cos θ−𝐴′′𝐴′′′ sin θ
𝑑𝑥′
= −𝜀 𝑥
𝑑𝑥
𝑑𝑥′ sin θ + 𝜀 𝑦
𝑑𝑦
𝑑𝑥′ cos θ − 𝛾𝑥𝑦
𝑑𝑦
𝑑𝑥′ sin θ
= − 𝜀 𝑥 − 𝜀 𝑦 sin θ cos θ − 𝛾𝑥𝑦sin2
θ
• By analogous reasoning,
𝛽 ≈ − 𝜀 𝑥 − 𝜀 𝑦 sin θ cos θ + 𝛾𝑥𝑦 cos2
θ

23. • Since, the shear strain 𝛾𝑥′𝑦′ of an angle included between the 𝑥′ 𝑦′
axes is 𝛽 + 𝛼,
𝛾𝑥′𝑦′ = −2 𝜀 𝑥 − 𝜀 𝑦 sin θ cos θ + 𝛾𝑥𝑦 cos2
θ − sin2
θ
or 𝛾𝑥′𝑦′ = − 𝜀 𝑥 − 𝜀 𝑦 sin 2θ + 𝛾𝑥𝑦 cos 2θ
• Both stresses and strains are second-rank tensors and mathematically
obey the same laws of transformation.

24. Mohr’s Circle for Two-Dimensional Strain
𝜀 𝑥′ =
𝜀 𝑥+𝜀 𝑦
2
+
𝜀 𝑥−𝜀 𝑦
2
cos 2θ +
𝛾 𝑥𝑦
2
sin 2θ
𝛾 𝑥′𝑦′
2
= −
𝜀 𝑥−𝜀 𝑦
2
sin 2θ +
𝛾 𝑥𝑦
2
cos 2θ
• In Mohr’s circle of strain, every point on the circle gives two values:
normal strain, and shear strain divided by 2. the vertical axis is
measured in terms of 𝛾 2.
• The center of the circle is at 𝜀 𝑥 + 𝜀 𝑦 2 , 0 , and the origin of
planes 𝐴 on the circle is at 𝜀 𝑥, 𝛾𝑥𝑦 2 .

25. Fig. 9: Mohr’s circle of strain using previous sign convention

26. • Conclusions:
1. The maximum normal stress is 𝜀1; the minimum is 𝜀2. These are the
principal strains, and no shear strains are associated with them. The
directions of the normal strains coincide with the directions of the
principal stresses.
𝜀 𝑥′ 𝑚𝑎𝑥
𝑚𝑖𝑛
= 𝜀1 𝑜𝑟 2 =
𝜀 𝑥+𝜀 𝑦
2
±
𝜀 𝑥−𝜀 𝑦
2
2
+
𝛾 𝑥𝑦
2
2
The planes on which the principal strains act can be defined as,
tan 2θ1 =
𝛾𝑥𝑦
𝜀 𝑥 − 𝜀 𝑦

27. 2. γmax is equal to twice the radius of the circle. Normal strains of
𝜀1 + 𝜀2 2 in two mutually perpendicular directions are associated
with the maximum shear strain.
3. The sum of normal strains in any two mutually perpendicular
directions is invariant, i.e., 𝜀1 + 𝜀2 = 𝜀 𝑥 + 𝜀 𝑦 = constant.
• Mohr’s strain circles degenerate to a point when two principal strains
are equal.
• For the stress in the 𝑧 direction,
𝜎𝑧 =
𝐸
1+𝜈 1−2𝜈
1 − 𝜈 𝜀 𝑧 + 𝜈 𝜀 𝑥 + 𝜀 𝑦
Then, since for plane stress 𝜎𝑧 = 0,
𝜀 𝑧 = −
𝜈
1+𝜈
𝜀 𝑥 + 𝜀 𝑦

28. Strain Rosettes
• Normal strains are measured along several closely clustered gage
lines, diagrammatically indicated in Fig. 10(a) by lines 𝑎 − 𝑎, 𝑏 − 𝑏,
and 𝑐 − 𝑐.
• Dividing the elongation by the gage length gives the strain in the θ1
direction, 𝜀θ1
.
• If the distances between the gage points are small, measurements
approximating the strains at a point are obtained.
• Arrangements of gage lines at a point in a cluster, Fig. 10, are called
strain rosettes.

29. Fig. 10: (a) General strain rosette; (b) rectangular or 45° strain rosette;
(c) equiangular or delta rosette.
• Rosettes usually consists of three single-element gages employed
together, as shown in Fig. 11. Metal-foil rosettes of this type are
available in a wide range of sizes, with active gage lengths varying from
0.8 to 12 mm.
• If angles θ1, θ2, and θ3 together with the corresponding strains 𝜀θ1
,
𝜀θ2
, and 𝜀θ3
are known from measurements,

30. 𝜀θ1
= 𝜀 𝑥 cos2θ1 + 𝜀 𝑦 sin2θ1 + 𝛾𝑥𝑦 sin θ1 cos θ1
𝜀θ2
= 𝜀 𝑥 cos2
θ2 + 𝜀 𝑦 sin2
θ2 + 𝛾𝑥𝑦 sin θ2 cos θ2
𝜀θ3
= 𝜀 𝑥 cos2
θ3 + 𝜀 𝑦 sin2
θ3 + 𝛾𝑥𝑦 sin θ3 cos θ3
• This set of equations can be solved for 𝜀 𝑥, 𝜀 𝑦, and 𝛾𝑥𝑦.
Fig. 11: Three-element metal-foil electrical-resistance strain gages
• To minimize computational work, the gages in a rosette are usually
arranged in a orderly manner.

31. • For example, in Fig. 10(b), θ1 = 0°, θ2 = 45°, and θ3 = 90°. The
arrangement of gage lines is known as the rectangular or the 45°
strain rosette.
𝜀 𝑥 = 𝜀0° 𝜀 𝑦 = 𝜀90°
2𝜀45° = 𝜀 𝑥 + 𝜀 𝑦 + 𝛾𝑥𝑦
• Another arrangement of gage lines is shown in Fig. 10(c). This is
known as equiangular, or the delta, or the 60° rosette.
𝜀 𝑥 = 𝜀0° 𝜀 𝑦 = 2𝜀60° + 2𝜀120° − 𝜀0° 3
and 𝛾𝑥𝑦 = 2 𝜀60° − 𝜀120° 3
• In rosettes with more than three lines, an additional gage line
measurement provides a check on the experimental work.

32. • For these rosettes, the invariance of the strains in the mutually
perpendicular directions can be used to check the data.
• The surface where the strains are measured is generally free of
significant normal surface stresses, i.e., 𝜎𝑧 = 0. therefore, this is a
plane stress problem.
𝜀1 =
𝜎1
𝐸
− 𝜈
𝜎2
𝐸
and 𝜀2 =
𝜎2
𝐸
− 𝜈
𝜎1
𝐸
• Solving these equations simultaneously for the principal stresses,
𝜎1 =
𝐸
1−𝜈2 𝜀1 + 𝜈𝜀2 𝜎2 =
𝐸
1−𝜈2 𝜀2 + 𝜈𝜀1

33. Part C- Yield and Fracture Criteria
• Some low-carbon steels below their transient temperatures of about
10°C (+ 50°F)become brittle, losing their ductile properties, and
behave like different materials, Fig. 12.
Fig. 12: Typical transition curve for stress or energy to fracture vs. temperature for carbon steel

34. • Most of the information on yielding and fracture of materials under
the action of biaxial stresses comes from experiments on thin-walled
cylinder.
• A typical arrangement for such an experiment is shown in Fig. 13.
Fig. 13: Arrangement for controlled ratios
of principal stresses

35. • By pressurizing the available space until the yielding or bursting
occurs, the elements of the wall are subjected to biaxial stresses of a
constant ratio 𝜎1 𝜎2 = 2.
• By applying an additional tensile force 𝑃 to the caps, the 𝜎2 stress is
increased to any predetermined amount 𝜎2 + 𝜎′′.
• By applying a compressive force, the 𝜎2 stress can be minimized or
eliminated.
• Analogous experiments with tubes simultaneously subjected to
torque, axial force, and pressure are used.

36. Maximum Shear-Stress Theory
• In a ductile material, slip occurs during yielding along critically
oriented planes.
• It is assumed that yielding of the material depends only on the
maximum shear stress that is attained within an element.
• If 𝜎𝑥 = ±𝜎1 ≠ 0, and 𝜎 𝑦 = 𝜏 𝑥𝑦 = 0,
𝜏 𝑚𝑎𝑥 ≡ 𝜏 𝑐𝑟 = ±
𝜎1
2
=
𝜎 𝑦𝑝
2
• In applying this criterion to a biaxial stress problem, two different
cases arise.
• In one case, the signs of the principal stresses 𝜎1 and 𝜎2 are the same.

37. • Taking them, to be tensile, Fig. 14(a) and setting 𝜎3 = 0, the resulting
Mohr’s principal stress structures are as shown in Fig. 14(b).
Fig. 14: Planes of
𝜏 𝑚𝑎𝑥 for
biaxial stresses

38. • Here the maximum shear stress is of the same magnitude as would occur
in a simple uniaxial stress, Figs. 14(a) and (c).
• Therefore, if 𝜎1 > 𝜎2 , 𝜎1 must not exceed 𝜎 𝑦𝑝. Similarly, if 𝜎2 >
𝜎1 , 𝜎2 must not be greater than 𝜎 𝑦𝑝.
𝜎1 ≤ 𝜎 𝑦𝑝 and 𝜎2 ≤ 𝜎 𝑦𝑝
• The second case is considered in Fig. 14(d)-(f), where the signs of 𝜎1 and
𝜎2 are opposite, and 𝜎3 = 0. the largest Mohr’s circle passes through 𝜎1
and 𝜎2, and the maximum shear stress,
𝜏 𝑚𝑎𝑥 = 𝜎1 + 𝜎2 2
• The alternative possible slip planes are identified in Fig. 14(d) and (f).
This maximum shear stress cannot exceed the shear yield criterion in
simple tension, i.e., 𝜏 𝑚𝑎𝑥 ≤ 𝜎 𝑦𝑝 2.

39. ±
𝜎1−𝜎2
2
≤
𝜎 𝑦𝑝
2
or, for impending yield,
𝜎1
𝜎 𝑦𝑝
−
𝜎2
𝜎 𝑦𝑝
= ±1
• A plot of this equation gives the two sloping lines shown in Fig. 15.
Fig. 15: Yield criterion based on maximum shear stress

40. • If a point defined by 𝜎1 𝜎 𝑦𝑝 and 𝜎2 𝜎 𝑦𝑝 falls on the hexagon shown
in Fig. 15, a material begins and continues to yield.
• No such stress points can lie outside the hexagon because one of the
three yield criteria equations given before for perfectly plastic
material would be violated.
• According to the maximum shear theory, if hydrostatic tensile or
compressive stresses are added, i.e., stresses such that 𝜎1
′
= 𝜎2
′
= 𝜎3
′
no change in the material response is predicted.
• Since the maximum shear stresses are defined on planes irrespective
of material directional properties, it is implicit that the material is
isotropic.
• The derived yield criterion for perfectly plastic material is often
referred to as the Tresca yield condition.

41. Maximum Distortion-Energy Theory
• Another widely accepted criterion of yielding for ductile isotropic
materials is based on energy concepts.
• The total elastic energy is divided into two parts: one associated with
the volumetric changes of the material, and the other causing shear
distortions.
• It is possible to consider the shear tensor of the three principal
stresses- 𝜎1, 𝜎2and 𝜎3- to consist of two additive component tensors.
• The elements of one component tensor are defined as the mean
“hydrostatic” stress:
𝜎 =
𝜎1+𝜎2+𝜎3
3

42. • The elements of the other tensor are 𝜎1 − 𝜎 , 𝜎2 − 𝜎 , and
𝜎3 − 𝜎 .
𝜎1 0 0
0 𝜎2 0
0 0 𝜎3
=
𝜎 0 0
0 𝜎 0
0 0 𝜎
+
𝜎1 − 𝜎 0 0
0 𝜎2 − 𝜎 0
0 0 𝜎3 − 𝜎
• This first stress tensor is called the spherical stress tensor or the
dilatational stress tensor.
• This resolution of the general state of stress is shown schematically in
Fig. 16.
• The sum of the stresses in Figs. 16(f) and (g) corresponds to the last
tensor of above matrix.
• The last tensor is called the deviatoric or distortional stress tensor.

43. • The strain energy per unit volume, i.e., strain density, for a three-
dimensional state of stress must be found. Using superposition,
𝑈0 = 𝑈𝑡𝑜𝑡𝑎𝑙 =
1
2
𝜎1 𝜀1 +
1
2
𝜎2 𝜀2 +
1
2
𝜎3 𝜀3
𝑈𝑡𝑜𝑡𝑎𝑙 =
1
2𝐸
𝜎1
2
+ 𝜎2
2
+ 𝜎3
2
−
𝜈
𝐸
𝜎1 𝜎2 + 𝜎2 𝜎3 + 𝜎3 𝜎1
• The strain energy per unit volume due to the dilatational stresses can
be determined from this equation by first setting 𝜎1 = 𝜎2 = 𝜎3 = 𝑝,
and then replacing 𝑝 by 𝜎 = 𝜎1 + 𝜎2 + 𝜎3 3.
𝑈 𝑑𝑖𝑙𝑎𝑡𝑎𝑡𝑖𝑜𝑛 =
3 1−2𝜈
2𝐸
𝑝2 =
1−2𝜈
6𝐸
𝜎1 + 𝜎2 + 𝜎3
2
𝑈 𝑑𝑖𝑠𝑡𝑜𝑟𝑡𝑖𝑜𝑛 =
1
12𝐺
𝜎1 − 𝜎2
2 + 𝜎2 − 𝜎3
2 + 𝜎3 − 𝜎1
2

44. • When one of the principal stresses reaches the yield point, 𝜎 𝑦𝑝, of the
material. The distortion strain energy for this is 2𝜎 𝑦𝑝
2 12𝐺.
• The basic law for yielding of an ideally plastic material,
𝜎1 − 𝜎2
2 + 𝜎2 − 𝜎3
2 + 𝜎3 − 𝜎1
2 = 2𝜎 𝑦𝑝
2
• For plane stress, 𝜎3 = 0,
𝜎1
𝜎 𝑦𝑝
2
−
𝜎1
𝜎 𝑦𝑝
𝜎2
𝜎 𝑦𝑝
+
𝜎2
𝜎 𝑦𝑝
2
= 1
• This is an equation of an ellipse, a plot of which is shown in Fig. 16.
• Any stress falling within the ellipse indicates that the material behaves
elastically. Points on the ellipse indicate that the material is yielding.

45. Fig. 16: Yield criterion based on maximum distortion energy
• In the three-dimensional stress space, the yield surface becomes a
cylinder with an axis having all three direction cosines equal to 1 3.
• Such a cylinder is shown in Fig. 17. the ellipse in Fig. 16 is simply the
intersection of this cylinder with the 𝜎1 − 𝜎2 plane.

46. • It can be shown also that the yield surface for the maximum shear
stress criterion is a hexagon that fits into the tube, Fig. 16.
Fig. 17: Yield surfaces for tri-axial state of stress
• This constitutive equation for perfectly plastic material is referred to as
the Huber-Hencky-Mises or the von Mises yield condition.

47. Comparison of Maximum-Shear and Distortion-
Energy Theories for Plane Stress
• A comparison between them for plane stress is shown in Fig. 18.
Fig. 18: Comparison of Tresca and von
Mises yield criteria
• If a stress point for the principal stresses 𝜎1 and 𝜎2 falls within these
curves, a material behaves elastically.

48. • The maximum shear-stress theory is more conservative.
• The yield criteria in the second and forth quadrant indicate smaller
strengths at yield than that for uniaxial stresses.
• The largest discrepancy occurs when two of the principal stresses are
equal but of opposite sign. This condition develops, for example, in
torsion of thin-walled tubes.
• According to the maximum shear-stress theory, when ±𝜎1 = ∓𝜎2,
these stresses at yield can reach only 𝜎 𝑦𝑝 2.
• The maximum distortion-energy theory limits this stress to
𝜎 𝑦𝑝 3 = 0.577𝜎 𝑦𝑝.

49. Maximum Normal Stress Theory
• Failure or fracture of a material occurs when the maximum normal
stress at a point reaches a critical value regardless of the other
stresses.
• The critical value of stress 𝜎 𝑢𝑙𝑡 is usually determined in a tensile
experiment, where the failure of a specimen is defined to be either
excessively large elongation or fracture.
• This theory applies well to brittle materials in all ranges of stresses,
providing a tensile principal stress exists.
• The maximum stress theory can be interpreted on graphs, Fig. 19.
• Failure occurs if points fall on the surface.

50. Comparison of Yield and Fracture Criteria
• Comparison of some classical experimental results with the yield and
fracture criteria is shown in Fig. 19.
Fig. 19: Comparison of yield
and fracture criteria with
test data

51. • The maximum normal stress theory appears to be best for brittle
materials and can be unsafe for ductile materials.
• If one of the principal stresses at a point is large in comparison with
the other, all theories give practically the same results.
• The discrepancy between the criteria is greatest in the second and
fourth quadrants, when both principal stresses are numerically equal.
• The improved model recognized the higher strengths of brittle
materials in biaxial compression than in tension.
• Therefore, the region in biaxial tension in the principal stress space is
made smaller than it is for biaxial compression, Fig. 20.
• In the second and fourth quadrant, a linear change between the two
of the above regions is assumed.

52. Fig. 20: Plausible fracture criteria Fig. 21: Dashed curve shows analytical fit for
for brittle materials three different strength concretes
• According to this theory, an existing crack will rapidly propagate if the
available elastic strain energy release rate is greater than the increase
in the surface energy of the crack.
• Careful recent experimental research on concrete specimens of
different strengths strongly corroborates this approach, Fig. 21.

53. Fig. 22: (a) Mohr envelopes, (b) failure planes at 𝐴 and 𝐴′,
(c) failure planes at 𝐵 and 𝐵′, (d) Mohr envelope solution in principal stress space.

54. • Another important attempt for rationalizing fracture of materials
having different properties in tension and compression is due to
Mohr.
• If the results of experiments in tension, compression, and shear are
available, the results can be represented on the same plot using their
respective largest principal stress circles, Fig. 22(a).
• The point of contact of the envelopes with the stress circles define
the state of stress at a fracture.
• The corresponding planes of stresses for points 𝐴 or 𝐴′ are shown in
Fig. 22(b), and a material such as duralumin does fracture in tension
at a flat angle as shown.
• By relating the fractures to either point 𝐵 or 𝐵′, the fracture occurs at
a steep angle characteristic of concrete cylinders tested in
compression, Fig. 22(c).

55. • The data from Fig. 22(a) can be replotted in the principal stress space
as in Fig. 22(d).
• Since in the first quadrant, the minimum principal stress 𝜎3 = 0, and
in the third quadrant, 𝜎3 = 0 is the maximum principal stress, per
Figs. 22(a)-(c), in these quadrants, the fracture lines in the principal
stress space are similar to those of Fig. 15.
• Part of the stress circles for tension and compression should be taken
as envelope ends.
• For a loose granular media such as sand, the straight-line Mohr
envelopes corresponds to the limiting condition of dry friction, 𝜇 =
tan ϕ, Fig. 23.
• Any circle tangent to the envelope, as at 𝐵, gives the state of critical
stress.

56. Fig. 23: Mohr envelopes for cohesion-less granular media
• If some cohesion can be developed by the media, the origin 𝑂 is
moved to the right such that at zero stress, the 𝜏 intercept is equal to
the cohesion.
• The fracture theory based on Mohr envelopes, using the largest
principal stress circles, neglects dependence on the intermediate
principal stress.