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

Method one. The first method is simply to substitute one of the θ s values into either
Equation (12.4) or Equation (12.6) and compute the corresponding shear stress. In addition
to giving the value of the maximum in-plane shear stress, this method has the advantage that
it directly associates a shear stress magnitude (including the proper sign) with the θ s angle.
Thus, given that shear stresses on orthogonal planes must be equal, the determination of the
stress for only one θ s angle is sufficient to define uniquely the shear stresses on both planes.
Since one is typically interested in finding both the principal stresses and the maximum
in-plane shear stress, an efficient computational approach for finding both the magnitude
and orientation of the maximum in-plane shear stress is as follows:
(a) From Equation (12.11), a specific value for θ p will be known.
(b) Depending on the sign of θ p and recognizing that θ p and θ s are always 45° apart, either
add or subtract 45° to find an orientation of a maximum in-plane shear stress plane θ s .
To obtain an angle θ s between +45° and −45° (inclusive), subtract 45° from a positive
value of θ p or add 45° to a negative value of θ p .
(c) Substitute the resulting value of θ s into either Equation (12.4) or Equation (12.6), and
compute the corresponding shear stress. The result is τ max , the maximum in-plane
shear stress.
(d) The result obtained from either Equation (12.4) or Equation (12.6) for θ s will furnish
both the magnitude and the sign of the maximum in-plane shear stress τ max . Obtaining
the sign is particularly valuable in this method because Method Two offers no direct
means for establishing the sign of τ max .
Method two. A general equation can be derived to give the magnitude of τ max by
substituting angle functions obtained from Equation (12.14) into Equation (12.6). The
result is
τ
max
which reduces to
σx − σ y ⎡ ± ( σx − σ y)/2
⎤ ⎡ ∓τ
xy
=− ⎢
⎥ + τ xy
2
2
⎢
2
⎢ ⎛ σx
− σ y⎞
σ σ
τ
τ
⎝
⎜
⎠
⎟ +
⎥ ⎢ ⎛ −
2
x y⎞
xy
⎣
⎢
⎦
⎥ ⎝
⎜
⎠
⎟ +
2 ⎣
⎢ 2
2
xy
⎤
⎥
⎥
⎦
⎥
503
PRINCIPAL STRESSES ANd
MAxIMuM SHEAR STRESS
τ
max
=±
⎛ σx
⎝
⎜
− σ
2 ⎠
2
y⎞
2
⎟ + τ xy
(12.15)
Note that Equation (12.15) has the same magnitude as the second term of Equation (12.12).
From Equation (12.15), the sign of τ max is ambiguous. The maximum shear stress differs
from the minimum shear stress only in sign. Unlike normal stress, which can be either
tension or compression, the sign of the maximum in-plane shear stress has no physical
significance for the material behavior of a stressed body. The sign simply indicates the direction
in which the shear stress acts on a particular plane surface.
A useful relation between the principal and the maximum in-plane shear stress is
obtained from Equations (12.12) and (12.15) by subtracting the values for the two inplane
principal stresses and substituting the value of the radical from Equation (12.15).
The result is
τ
max
σ
=
− σ
p1 p2
2
(12.16)