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

Table 13.2 Absolute maximum Shear Strains
(a) If both ε p1 and ε p2 are positive, then
γ = ε − ε = ε − 0 = ε
absmax p1 p3 p1 p1
Principal Strain Element
ε p3 =ε z = 0
Absolute Maximum Shear Strain Element
π
2
+γ abs max
ε p1
π
2
–
γ abs max
(b) If both ε p1 and ε p2 are negative, then
γ = ε − ε = 0 − ε =−ε
absmax p3 p2 p2 p2
ε p3 = ε z = 0
π
2
+γ abs max
ε p2
π
2
–
γ abs max
(c) If ε p1 is positive and ε p2 is negative, then
γ = ε − ε
absmax p1 p2
ε p2
π
2
+γ abs max
ε p1
π
2
–
γ abs max
These conditions apply only to a state of plane strain. As will be shown in Sections
13.7 and 13.8, the third principal strain will not be zero for a state of plane stress.
13.5 presentation of Strain Transformation Results
Principal strain and maximum in-plane shear strain results should be presented with a
sketch that depicts the orientation of all strains. Strain results can be conveniently shown
on a single element.
Draw an element rotated at the angle θ p calculated from Equation (13.9), which will
give a value between +45° and −45° (inclusive):
tan2θ
p
γ xy
=
ε − ε
x
y
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