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

• When θ p is positive, the element is rotated in a counterclockwise sense from the
reference x axis. When θ p is negative, the rotation is clockwise.
• Note that the angle calculated from Equation (13.9) does not necessarily give the orientation
of the ε p1 direction. Either ε p1 or ε p2 may act in the θ p direction given by that equation.
The principal strain oriented at θ p can be determined from the following two-part rule:
• If ε x − ε y is positive, then θ p indicates the orientation of ε p1 .
• If ε x − ε y is negative, then θ p indicates the orientation of ε p2 .
• Elongate or contract the element into a rectangle according to the principal strains acting
in the two orthogonal directions. If a principal strain is positive, then the element is
elongated in that direction. The element is contracted if the principal strain is negative.
• Add either tension or compression arrows, labeled with the corresponding strain
magnitudes, to each edge of the element.
• To show the distortion caused by the shear strain, draw a diamond shape inside of the
rectangular principal strain element. The corners of the diamond should be located at
the midpoint of each edge of the rectangle.
• The maximum in-plane shear strain calculated from Equation (13.11) or Equation
(13.12) will be a positive value. Since a positive shear strain causes the angle between
two axes to decrease, label one of the acute angles with the value π/2 − γ max .
ExAmpLE 13.2
The strain components at a point in a body subjected to plane strain are ε x =
−680 µε, ε y = 320 µε, and γ xy = −980 m µrad. The deflected shape of an element
that is subjected to these strains is shown. Determine the principal strains, the
maximum in-plane shear strain, and the absolute maximum shear strain at point O.
Show the principal strain deformations and the maximum in-plane shear strain
distortion in a sketch.
ε y dy
dy
y
π
– γ xy
2
SolutioN
From Equation (13.10), the in-plane principal strains are
ε
p1, p2
2 2
εx + εy ⎛εx − εy⎞
γ xy
= ± ⎜ ⎟ + ⎛ 2 ⎝ 2 ⎠ ⎝ ⎜
⎞
⎟
2 ⎠
O
dx
ε x dx
x
680 320
= − + ±
2
=− 180 ± 700
2 2
⎛−680 − 320 ⎞ 980
⎜ ⎟ + ⎛−
⎞
⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
= 520 µε, − 880 µε Ans.
and from Equation (13.11), the maximum in-plane shear strain is
2 2
γ max ⎛ εx − εy⎞
γ xy
=±
2 ⎝
⎜
2 ⎠
⎟ + ⎛ ⎝ ⎜ ⎞
2 ⎠
⎟
2 2
⎛ −680 − 320⎞
980
=±
⎝
⎜
2 ⎠
⎟ + ⎛ − ⎞
⎝
⎜
2 ⎠
⎟
= 700 µ rad
∴ γ = 1, 400 µ rad
Ans.
max
549