strength of material complex stress

1. Chapter nine
COMPLEX STRESS

2. Introduction
From the previous analysis the determination of stress distributions produced separately by
axial load, bending moment, shear force and torsion. However, in many practical situations
some or all of these force systems act simultaneously so that the various stresses are combined
to form complex systems which may include both direct and shear stresses. In such cases it is
no longer a simple matter to predict the mode of failure of a structural member. Therefore in
this chapter the stress and strain subjected to complex loading will be examined.
Plane stress
The stress conditions that, when analyzing bars in tension and compression, shafts in torsion,
and beams in bending are examples of a state of stress called plane stress. Consider an
infinitesimal element,

3. When the material is in plane stress in the xy plane, only the x and y faces of the element are
subjected to stresses, and all stresses act parallel to the x and y axis (with 𝜎𝑍, 𝜏𝑍𝑋, 𝜏𝑍𝑌 = 0) and is
defined by the stress components (𝜎𝑥, 𝜎𝑦, 𝜏𝑋𝑌). The stress components (𝜎𝑥1, 𝜎𝑦1, 𝜏𝑋1𝑌1) associated
with the element are determined after it has been rotated through an angle θ about the z axes. These
components are given in terms of 𝜎𝑥, 𝜎𝑦, 𝜏𝑋𝑌 and 𝜃
A normal stress a has a subscript that identifies the face on which the stress acts; for instance, the stress 𝜎𝑥acts on the x face of the
element and the stress 𝜎𝑦 acts on the y face of the element. The sign convention for normal stresses tension is positive and
compression is negative.
A shear stress has two subscripts—the first subscript denotes the face on which the stress acts, and the second gives the direction on
that face.
This sign convention for shear stresses is easy to remember, shear stress is positive when the directions associated with its
subscripts are plus-plus or minus-minus; the shear stress is negative when the directions are plus-minus or minus-plus.
If the area of the oblique face is∆𝐴, the areas of the vertical and horizontal faces are equal to
∆𝐴 cos 𝜃 and ∆𝐴 sin 𝜃, respectively.

4.  It follows that the forces exerted on the three faces are as shown in Fig. 7.6b.
 Using components along the x’ and y’ axes, we write the following equilibrium equations:
 Solving the first equation for 𝜎𝑥′ and the second for 𝜏𝑥′𝑦′ we have
 Recalling the trigonometric relations
sin 2𝜃= 2sin 𝜃 cos 𝜃

5. 𝜎𝑥′ =
𝜎𝑥+𝜎𝑦
2
+ [
𝜎𝑥−𝜎𝑦
2
]cos 2𝜃 + 𝜏𝑥𝑦 sin 2𝜃
𝜏𝑥𝑦′ = −
𝜎𝑥−𝜎𝑦
2
sin 2𝜃 + 𝜏𝑥𝑦 cos 2𝜃
Special cases of plane stress
 Uniaxial stress: if all stresses acting on xy element are zero except for normal stress 𝜎𝑥.
𝜎𝑥′ =
𝜎𝑥
2
[1 + cos 2𝜃]
𝜏𝑥𝑦′ = −
𝜎𝑥
2
sin 2𝜃
 Pure shear stress: when the element is subjected to shear stress.
𝜎𝑥′ = 𝜏𝑥𝑦 sin 2𝜃
𝜏𝑥𝑦′ = 𝜏𝑥𝑦 cos 2𝜃
 Biaxial stress: in which the xy element is subjected to normal stress in both x and y directions but with
out shear. Case for biaxial stress is the thin walled pressure vessels.
𝜎𝑥′ =
𝜎𝑥 + 𝜎𝑦
2
+ [
𝜎𝑥 − 𝜎𝑦
2
]cos 2𝜃
𝜏𝑥𝑦′ = −[
𝜎𝑥 − 𝜎𝑦
2
] sin 2𝜃

8. Principal stress and maximum shear stress
The transformation equations for plane stress show that the normal stresses(𝜎𝑥′) and the shear stresses(𝜏𝑥′𝑦′) vary
continuously as the axes are rotated through the angle 𝜃. Both the normal and shear stresses reach maximum and minimum
values at 90° intervals. Not surprisingly, these maximum and minimum values are usually needed for design purposes.
 Principal stress: The maximum and minimum normal stresses, called the principal stresses, can be found from the
transformation equation for the normal stress (𝜎𝑥′). By taking the derivative of (𝜎𝑥′) with respect to 𝜃 and setting it equal
to zero. we obtain an equation from which we can find the values of 𝜃 at which 𝜎𝑥′ is a maximum or minimum. The
equation for the derivative is
from which we get
 The subscript p indicates that the angle 𝜃p defines the orientation of the principal planes, that is, the planes on which the
principal stresses act. the angle 𝜃p has two values that differ by 90°, one value between 0 and 90° and the other between
90° and 180°. The two values of 𝜃p are known as the principal angles.

9. We can also obtain general formulas for the principal stresses.
 The quantity R is always a positive number and, like the other two sides of the triangle, has units of stress.
From the triangle we obtain two additional relations:
 Now we substitute these expressions for cos 2𝜃p, and sin 2𝜃p into Eq. (7-4a) and obtain the algebraically
larger of the two principal stresses, denoted by 𝜎1 :
 The smaller of the principal stresses, denoted by 𝜎2, may be found from the condition that the sum of the
normal stresses on perpendicular planes is constant

10.  The preceding formulas for 𝜎1 and 𝜎2 can be combined into a single formula for the principal stresses:
 Maximum Shear Stresses: The shear stresses acting on inclined planes are given by the second transformation
equation, taking the derivative of this, with respect to 𝜃 and setting it 'equal to zero.
 The subscript s indicates that the angle 𝜃s, defines the orientation of the planes of maximum positive and
negative shear stresses.
 Comparing Eq. for 𝜃s, with Eq. for 𝜃p shows that
 From this equation we can obtain a relationship between the angles 𝜃s, and 𝜃s.
 This equation shows that the planes of maximum shear stress occur at 45° to the principal planes

11.  the corresponding maximum shear stress is obtained by substituting the expressions for cos 2𝜃s1 and sin
2𝜃s1, into the second transformation equation
 The maximum negative shear stress 𝜏𝑚𝑖𝑛 has the same magnitude but opposite sign.