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

ExAmpLE 12.5
y
9 ksi
7 ksi
x
11 ksi
Consider a point in a structural member that is subjected to plane stress. Normal and shear
stresses acting on horizontal and vertical planes at the point are shown.
(a) Determine the principal stresses and the maximum in-plane shear stress acting at the
point.
(b) Show these stresses in an appropriate sketch.
(c) Determine the absolute maximum shear stress at the point.
Plan the Solution
The stress transformation equations derived in the preceding section will be used to compute
the principal stresses and the maximum shear stress acting at the point.
SolutioN
(a) From the given stresses, the values to be used in the stress transformation equations
are σ x = 11 ksi, σ y = −9 ksi, and τ xy = −7 ksi. The in-plane principal stress magnitudes
can be calculated from Equation (12.12):
σ
p1, p2
σx + σ y ⎛ σx − σ y⎞
= ±
τ
⎝
⎜
⎠
⎟ +
2 2
(11ksi) + ( −9ksi)
=
±
2
= 13.21 ksi, −11.21 ksi
2
2
xy
⎛ (11ksi) − ( −9ksi)
⎞
⎝
⎜
⎠
⎟
2
2
+ ( −7 ksi)
2
Therefore, we have the following:
σ
σ
p1
p2
= 13.21 ksi = 13.21 ksi (T)
=− 11.21 ksi = 11.21 ksi (C)
The maximum in-plane shear stress can be computed from Equation (12.15):
τ
max
=±
⎛ σx
− σ
⎝
⎜
2 ⎠
=± 12.21 ksi
2
y⎞
2
⎟ + τ xy = ±
⎛ (11ksi) − ( −9ksi)
⎞
⎝
⎜
2 ⎠
⎟
2
+ ( −7 ksi)
2
On the planes of maximum in-plane shear stress, the normal stress is simply the
average normal stress, as given by Equation (12.17):
σ avg
σx
+ σ
=
2
y
11 ksi + ( −9ksi)
=
= 1ksi = 1ksi (T)
2
(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( −7ksi)
= =
σ − σ 11 ksi − ( −9ksi)
x
y
∴ θ = − 17.5°
p
14
= −
20
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