Page 254 - Shigley's Mechanical Engineering Design

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Failures Resulting from Static Loading 229
Figure 5–13 Coulomb-Mohr
failure line
Mohr’s largest circle for a
general state of stress. B
3 B 2
B
1
O
–S c C 3 C 2 C 1 S t
3
1

where B 1C 1 = S t /2, B 2 C 2 = (σ 1 − σ 3 )/2, and B 3C 3 = S c /2, are the radii of the right,
center, and left circles, respectively. The distance from the origin to C 1 is S t /2, to C 3 is
S c /2, and to C 2 (in the positive σ direction) is (σ 1 + σ 3 )/2. Thus
σ 1 − σ 3 S t S c S t
− −
2 2 2 2
=
S t σ 1 + σ 3 S t S c
− +
2 2 2 2
Canceling the 2 in each term, cross-multiplying, and simplifying reduces this equa-
tion to
σ 1 σ 3
− = 1 (5–22)
S t S c
where either yield strength or ultimate strength can be used.
For plane stress, when the two nonzero principal stresses are σ A ≥ σ B , we have
a situation similar to the three cases given for the MSS theory, Eqs. (5–4) to (5–6).
That is, the failure conditions are
Case 1: σ A ≥ σ B ≥ 0. For this case, σ 1 = σ A and σ 3 = 0. Equation (5–22)
reduces to
(5–23)
σ A ≥ S t
Case 2: σ A ≥ 0 ≥ σ B . Here, σ 1 = σ A and σ 3 = σ B , and Eq. (5–22) becomes

σ A σ B
− ≥ 1 (5–24)
S t S c
Case 3: 0 ≥ σ A ≥ σ B . For this case, σ 1 = 0 and σ 3 = σ B , and Eq. (5–22) gives
(5–25)
σ B ≤−S c
A plot of these cases, together with the normally unused cases corresponding to
σ B ≥ σ A , is shown in Fig. 5–14.
For design equations, incorporating the factor of safety n, divide all strengths by n.
For example, Eq. (5–22) as a design equation can be written as

σ 1 σ 3 1
− = (5–26)
S t S c n