Page 245 - Shigley's Mechanical Engineering Design

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220 Mechanical Engineering Design
beginning of yield, and when loaded to fracture, fracture lines are also seen at angles
approximately 45° with the axis of tension. Since the shear stress is maximum at 45°
from the axis of tension, it makes sense to think that this is the mechanism of failure. It
will be shown in the next section, that there is a little more going on than this. However,
it turns out the MSS theory is an acceptable but conservative predictor of failure; and
since engineers are conservative by nature, it is quite often used.
Recall that for simple tensile stress, σ = P/A, and the maximum shear stress
occurs on a surface 45° from the tensile surface with a magnitude of τ max = σ/2. So the
maximum shear stress at yield is τ max = S y /2. For a general state of stress, three prin-
cipal stresses can be determined and ordered such that σ 1 ≥ σ 2 ≥ σ 3 . The maximum
shear stress is then τ max = (σ 1 − σ 3 )/2 (see Fig. 3–12). Thus, for a general state of
stress, the maximum-shear-stress theory predicts yielding when
σ 1 − σ 3 S y
τ max = ≥ or σ 1 − σ 3 ≥ S y (5–1)
2 2
Note that this implies that the yield strength in shear is given by
(5–2)
S sy = 0.5S y
which, as we will see later is about 15 percent low (conservative).
For design purposes, Eq. (5–1) can be modified to incorporate a factor of safety, n.
Thus,
S y S y
τ max = or σ 1 − σ 3 = (5–3)
2n n
Plane stress is a very common state of stress in design. However, it is extremely
important to realize that plane stress is a three-dimensional state of stress. Plane stress
transformations in Sec. 3–6 are restricted to the in-plane stresses only, where the in-
plane principal stresses are given by Eq. (3–13) and labeled as σ 1 and σ 2 . It is true that
these are the principal stresses in the plane of analysis, but out of plane there is a third
principal stress and it is always zero for plane stress. This means that if we are going to
use the convention of ordering σ 1 ≥ σ 2 ≥ σ 3 for three-dimensional analysis, upon
which Eq. (5–1) is based, we cannot arbitrarily call the in-plane principal stresses σ 1
and σ 2 until we relate them with the third principal stress of zero. To illustrate the MSS
theory graphically for plane stress, we will first label the principal stresses given by
Eq. (3–13) as σ A and σ B , and then order them with the zero principal stress according
to the convention σ 1 ≥ σ 2 ≥ σ 3 . Assuming that σ A ≥ σ B , there are three cases to con-
sider when using Eq. (5–1) for plane stress:

Case 1: σ A ≥ σ B ≥ 0. For this case, σ 1 = σ A and σ 3 = 0. Equation (5–1)
reduces to a yield condition of
(5–4)
σ A ≥ S y
Case 2: σ A ≥ 0 ≥ σ B . Here, σ 1 = σ A and σ 3 = σ B , and Eq. (5–1) becomes
(5–5)
σ A − σ B ≥ S y
Case 3: 0 ≥ σ A ≥ σ B . For this case, σ 1 = 0 and σ 3 = σ B , and Eq. (5–1) gives
(5–6)
σ B ≤−S y
Equations (5–4) to (5–6) are represented in Fig. 5–7 by the three lines indicated in the
σ A ,σ B plane. The remaining unmarked lines are cases for σ B ≥ σ A , which completes
the stress yield envelope but are not normally used. The maximum-shear-stress theory
predicts yield if a stress state is outside the shaded region bordered by the stress yield
envelope. In Fig. 5–7, suppose point a represents the stress state of a critical stress element