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

(b) The principal stresses and the maximum in-plane shear stress must be shown in an
appropriate sketch. The angle θ p indicates the orientation of one principal plane relative
to the reference x face. From Equation (12.11),
tan2θ
p
2τ
xy 2( −55 MPa)
= =
= − 110
σ − σ 70 MPa − 150 MPa −80
x
y
∴ θ = 27.0°
p
The angle θ p is positive; consequently, the angle is turned
counterclockwise from the x axis. Since σ 110 MPa
x − σ y is negative,
θ p indicates the orientation of σ p2 = 42.0 MPa. The
150 MPa
other principal stress, σ
68 MPa
p1 = 178.0 MPa, acts on a perpendicular
plane. The in-plane principal stresses are shown 55 MPa
in the accompanying figure.
70 MPa
The maximum in-plane shear stress and the associated
average normal stress are shown on the sloped face
x
27.0°
of the wedge, which is rotated 45° from the principal
planes. Note that the arrow for τ max starts on the σ p1 side
of the wedge and points toward the σ p2 side.
(c) Since σ p1 and σ p2 are both positive values, the absolute maximum shear stress will be
greater than the maximum in-plane shear stress. In this example, the three principal
stresses are σ p1 = 178 MPa, σ p2 = 42 MPa, and σ p3 = 0. The maximum principal stress
is σ max = 178 MPa, and the minimum principal stress is σ min = 0. The absolute maximum
shear stress can be computed from Equation (12.18):
τ
absmax
σ
=
max
− σ
2
min
178 MPa − 0
=
2
= 89.0 MPa
Ans.
The absolute maximum shear stress acts on a plane whose normal does not lie in the
x–y plane.
178 MPa
42 MPa
σ p2 side of
the wedge
σp1 side of
the wedge
Results sketched by using
the wedge element format.
mecmovies
ExAmpLE
m12.9 Stress transformation learning tool
Illustrates the correct usage of the stress transformation
equations in determining stresses acting on a specified plane,
principal stresses, and the maximum in-plane shear stress
state for stress values specified by the user.
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