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

660
COMbINEd LOAdS
from which it follows that
1
u
v
2 E
[ 2 2 2
= σ p1
+ σ p2
+ σ p3
− 2 ( σ p1σ p2 + σ p2σ p3 + σ p3σ
p1
)] (15.2)
The total strain energy can be resolved into components associated with a volume
change (u v ) and a distortion (u d ) by considering the principal stresses to be made up of two
sets of stresses as indicated in Figures 15.9a–c. The state of stress depicted in Figure 15.9c
will produce only distortion (no volume change) if the sum of the other three normal strains
is zero—that is, if
E( ε + ε + ε ) = [( σ − p) − v( σ + σ − 2 p)]
p1 p2 p3 d p1 p2 p3
+ [( σ − p) − v( σ + σ − 2 p)]
p2 p3 p1
+ [( σ − p) − v( σ + σ − 2 p)] = 0
p3 p1 p2
where p is the hydrostatic stress. This equation reduces to
Therefore, the hydrostatic stress is
(1 − 2 v)( σ + σ + σ − 3 p) = 0
p1 p2 p3
1
p =
3 ( σ p 1 + σ p 2 + σ p 3)
From Equation (13.16), the three normal strains due to the hydrostatic stress p are
1
ε v = (1 − 2 vp )
E
and the energy resulting from the hydrostatic stress (i.e., the volume change) is
u
v
pε
v
= 3
⎛ ⎞
⎝ 2 ⎠ =
3
2
1−
2 v
2
(1 − 2 v)
p = ( σ p + σ p + σ p )
E 6E
1 2 3 2
u = u − u
d
The energy resulting from the distortion (i.e., the change in shape) is
v
1
2 2 2
= ⎡3( σ p1
+ σ p2
+ σ p3) − 6 v( σ p1σ p2 + σ p2σ p3 + σ p3σ p1) − (1 − 2 v)( σ p1 + σ p2 + σ p3)
2
6E
⎣
⎤
⎦
σ
p2
p
σp2− p
= +
σ
p3
σp1
p
p
σp3− p
(a) (b) (c)
FIGURE 15.9 Expressing the state of stress in terms of volume-change components and
distortion components.
σp1− p