Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

In this figure, we will assume that τ xy and (σ x − σ y ) are both positive or both negative
quantities. From trigonometry, expressions can be developed for sin 2θ p and cos 2θ p , two
terms that are needed for the solution of Equation (12.5):
sin2θ
p
=
τ
xy
⎛ σx
− σ y⎞
τ
⎝
⎜
⎠
⎟ +
2
2
2
xy
cos2θ
p
=
( σ − σ )/2
x
⎛ σx
− σ y⎞
τ
⎝
⎜
⎠
⎟ +
2
y
2
2
xy
501
PRINCIPAL STRESSES ANd
MAxIMuM SHEAR STRESS
When these functions of 2θ p are substituted into Equation (12.5) and simplified, one obtains
σ
p1
σx + σ y ⎛ σx − σ y⎞
= +
τ
2 ⎝
⎜
2 ⎠
⎟ +
2
2
xy
A similar expression is obtained for σ p2 by repeating these steps with the principal angle
2θ p + 180°:
σ
p2
σx + σ y ⎛ σx − σ y⎞
= −
τ
2 ⎝
⎜
2 ⎠
⎟ +
2
2
xy
These two equations can then be combined into a single equation for the two in-plane principal
stresses σ p1 and σ p2 :
σ
p1, p2
2
σx + σ y ⎛ σx − σ y⎞
2
= ±
τ xy
2 ⎝
⎜
2 ⎠
⎟ + (12.12)
Equation (12.12) does not directly indicate which principal stress, either σ p1 or σ p2 , is associated
with each principal angle, and that is an important consideration. The solution of Equation
(12.11) always gives a value of θ p between +45° and −45° (inclusive). The principal
stress associated with this value of θ p can be determined from the following two-part rule:
• If the term σ x − σ y is positive, then θ p indicates the orientation of σ p1 .
• If the term σ x − σ y is negative, then θ p indicates the orientation of σ p2 .
The other principal stress is oriented perpendicular to θ p .
The principal stresses determined from Equation (12.12) may both be positive, may
both be negative, or may be of opposite signs. In naming the principal stresses, σ p1 is the more
positive value algebraically (i.e., the algebraically larger value). If one or both of the principal
stresses from Equation (12.12) are negative, σ p1 can have a smaller absolute value than σ p2 .
Shear Stresses on principal planes
As shown in the previous discussion, the values of θ p that are solutions of Equation (12.11)
will produce values of τ nt = 0 in Equation (12.6). Therefore, the shear stress on a principal
plane must be zero. This is a very important conclusion.
This characteristic of principal planes can be restated in the following manner:
If a plane is a principal plane, then the shear stress
acting on the plane must be zero.
The converse of this statement is also true:
If the shear stress on a plane is zero, then that
plane must be a principal plane.