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

y
y
π
2
γ xy
x
FIGURE 13.6a Positive shear
strain γ xy at origin.
–
π
2
FIGURE 13.6b Negative
shear strain γ xy at origin.
– γ xy
x
allow the determination of the shear strain γ nt associated with any two orthogonal lines
oriented in the n and t directions in the x–y plane. With these equations, the sign conventions
used in their development must be rigorously followed:
1. Normal strains that cause elongation are positive, and strains that cause contraction are
negative.
2. A positive shear strain decreases the angle between the two lines at the origin of coordinates
(Figure 13.6a). A negative shear strain increases the angle between the two lines
(Figure 13.6b).
3. Angles measured counterclockwise from the reference x axis are positive. Conversely,
angles measured clockwise from the reference x axis are negative.
4. The n–t–z axes have the same order as the x–y–z axes. Both sets of axes form a righthand
coordinate system.
ExAmpLE 13.1
An element of material at point O is subjected to a state of plane strain with the
strains specified as ε x = 600 µε, ε y = −300 µε, and γ xy = 400 µrad. The deflected
shape of the element subjected to these strains is shown. Determine the strains
acting at point O on an element that is rotated 40° counterclockwise from the
original position.
ε y dy
y
Plan the Solution
The strain transformation equations will be used to compute ε n , ε t , and γ nt .
SolutioN
The strain transformation equation
dy
O
π
– γ xy
2
dx
x
2 2
ε = ε cos θ + ε sin θ + γ sinθcosθ
(13.3)
n x y xy
ε x dx
will be used to compute the normal strains ε n and ε t . Since counterclockwise
angles are positive, the angle to be used in this instance is θ = 40°. First,
y
2 2
ε = (600)cos (40 ° ) + ( − 300)sin (40 ° ) + (400)sin(40 ° )cos(40 ° )
n
= 425 µε
Next, to compute the normal strain ε t , use an angle of θ = 40° + 90° = 130° in
Equation (13.3):
t
n
2 2
ε = (600)cos (130 ° ) + ( − 300)sin (130 ° ) + (400)sin(130 ° )cos(130 ° )
t
=− 125 µε
O
40°
x
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