Page 247 - Shigley's Mechanical Engineering Design

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222 Mechanical Engineering Design
–
2 av 2 av

= +
1 av 1 – av

3 –
3 av av
> > 3
2
1
(a) Triaxial stresses (b) Hydrostatic component (c) Distortional component
Figure 5–8
(a) Element with triaxial stresses; this element undergoes both volume
change and angular distortion. (b) Element under hydrostatic normal
stresses undergoes only volume change. (c) Element has angular
distortion without volume change.

1
The strain energy per unit volume for simple tension is u = σ. For the element
2
1
of Fig. 5–8a the strain energy per unit volume is u = [ 1 σ 1 + 2 σ 2 + 3 σ 3 ].
2
Substituting Eq. (3–19) for the principal strains gives
1 2 2 2
u = σ + σ + σ − 2ν(σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 ) (b)
1
2
3
2E
The strain energy for producing only volume change u v can be obtained by substitut-
ing σ av for σ 1 , σ 2 , and σ 3 in Eq. (b). The result is
2
3σ av
u v = (1 − 2ν) (c)
2E
If we now substitute the square of Eq. (a) in Eq. (c) and simplify the expression, we get
1 − 2ν 2 2 2
u v = σ + σ + σ + 2σ 1 σ 2 + 2σ 2 σ 3 + 2σ 3 σ 1 (5–7)
1
3
2
6E
Then the distortion energy is obtained by subtracting Eq. (5–7) from Eq. (b). This
gives
2
2
1 + ν (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) 2
u d = u − u v = (5–8)
3E 2
Note that the distortion energy is zero if σ 1 = σ 2 = σ 3 .
For the simple tensile test, at yield, σ 1 = S y and σ 2 = σ 3 = 0, and from Eq. (5–8)
the distortion energy is
1 + ν 2
u d = S y (5–9)
3E
So for the general state of stress given by Eq. (5–8), yield is predicted if Eq. (5–8)
equals or exceeds Eq. (5–9). This gives
2 2 2 1/2
(σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
≥ S y (5–10)
2
If we had a simple case of tension σ, then yield would occur when σ ≥ S y . Thus, the
left of Eq. (5–10) can be thought of as a single, equivalent, or effective stress for the
entire general state of stress given by σ 1 ,σ 2 , and σ 3 . This effective stress is usually