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

(b) Principal and Maximum in-Plane Shear Strains
The normal strains in the x, y, and z directions can be computed from Equations (13.24):
1 1
( )
−6
εx = σx − νσ y = [70 MPa − (0.3125)(81.2 MPa)] = 212.5 × 10 mm/mm
E
210,000 MPa
1 1
( )
−6
εy = σ y − νσ x = [81.2 MPa − (0.3125)(70 MPa)] = 282.5 × 10 mm/mm
E
210,000 MPa
ν 0.3125
6
ε z ( σ x σ
−
=− + y)
= − [81.2 MPa + 70MPa] =− 225 × 10 mm/mm
E
210,000 MPa
y
282.5 με
π
−70 μrad
2
212.5 με
Since γ xy = 0, the strains ε x and ε y are also the principal strains. Why? We know that
there is never a shear strain associated with the principal strain directions. Conversely,
we can also conclude that directions in which the shear strain is zero must
also be principal strain directions. Therefore,
ε = 282.5 µε ε = 212.5 µε ε =− 225 µε Ans.
p1 p2 p3
From Equation (13.12), the maximum in-plane shear strain can be determined from
ε p1 and ε p2 :
γmax = εp1 − ε p2
= 282.5 − 212.5 = 70 µ rad
Ans.
x
The in-plane principal strain deformations and the maximum in-plane shear strain
distortion are shown in the sketch.
(c) Absolute Maximum Shear Strain
To determine the absolute maximum shear strain, three possibilities
must be considered (see Table 13.2):
γ
2
y–z plane
γ = ε − ε
(i)
absmax p1 p2
γ = ε − ε
(ii)
absmax p1 p3
γ = ε − ε
(iii)
absmax p2 p3
P 3
(–225, 0)
x–z plane
R = 253.75
x–y plane These possibilities can be readily visualized with Mohr’s circle
for strains. The combinations of ε and γ that are possible in the
P 2 (282.5, 0) x–y plane are shown by the small circle between point P 1 (which
(212.5, 0) corresponds to the y direction) and point P 2 (which represents
P ε 1
the x direction). The radius of this circle is relatively small;
therefore, the maximum shear strain in the x–y plane is small
(γ max = 70 µrad). The steel plate in this problem is subjected to
plane stress, and consequently, the normal stress σ p3 = σ z = 0.
However, the normal strain in the z direction will not be zero.
For this problem, ε p3 = ε z = −225 µε. When this principal strain
is plotted on Mohr’s circle (i.e., point P 3 ), it becomes evident
that the out-of-plane shear strains will be much larger than the
shear strain in the x–y plane.
The largest shear strain will thus occur in an out-of-plane direction—in this instance,
a distortion in the y–z plane. Accordingly, the absolute maximum shear strain will be
γ = ε − ε = 282.5 − ( − 225) = 507.5 µ rad
Ans.
absmax p1 p3
570