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

t
n
Finally, to compute the shear strain γ nt from Equation (13.5), use an angle
of θ = 40°:
εt dt
dt
π
– γ nt
2
O
dn
εn dn
2 2
γ =−2[600 − ( − 300)]sin(40 ° )cos(40 ° ) + (400)[cos (40 ° ) − sin (40 ° )]
nt
=− 817 µ rad
The computed strains tend to distort the element as shown. The positive
normal strain ε n means that the element elongates in the n direction. The
negative value for ε t indicates that the element contracts in the t direction.
Although it initially seems counterintuitive, note that the negative shear
strain γ nt = −817 µrad means that the angle between the n and t axes actually
becomes greater than 90° at point O.
mecmovies
ExAmpLES
m13.1 The thin rectangular plate is uniformly deformed
such that ε x = −700 µε, ε y = −500 µε, and γ xy = +900 µrad.
Determine the normal strain
(a) along diagonal AC.
(b) along diagonal BD.
m13.3 A thin triangular plate is uniformly deformed
such that, after deformation, the edges of the triangle are
measured as AB = 300.30 mm, BC = 299.70 mm, and
AC = 360.45 mm. Determine the strains ε x , ε y , and γ xy in
the plate.
m13.2 The thin rectangular plate is subjected to strains
ε x = +900 µε, ε y = −600 µε, and γ xy = −850 µrad. Determine
the normal strains ε n and ε t , and the shear strain γ nt , for
θ = +50°.
Before deformation
After deformation
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