Page 251 - Shigley's Mechanical Engineering Design

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226 Mechanical Engineering Design
(b) From Eq. (3–13), the nonzero principal stresses are

2
60 + 40 60 − 40
2
σ A ,σ B = ± + (−15) = 68.0, 32.0 kpsi
2 2
The ordered principal stresses are σ A = σ 1 = 68.0,σ B = σ 2 = 32.0,σ 3 = 0 kpsi.
2
DE σ = 68 − 68(32) + 68 2 1/2 = 59.0 kpsi

Answer n = S y = 100 = 1.70
σ 59.0
MSS Noting that the two nonzero principal stresses are both positive, τ max will be
from the largest Mohr’s circle which will incorporate the third principle stress at zero.
From Eq. (3–16),
68.0 − 0
σ 1 − σ 3
τ max = = = 34.0 kpsi
2 2
S y /2 100/2
Answer n = = = 1.47
τ max 34.0
(c) This time, we shall obtain the factors of safety directly from the xy components
of stress.
DE From Eq. (5–15),
2
2
2
2

σ = (σ − σ x σ y + σ + 3τ ) 1/2 = (40 + 3(45) 2 1/2 = 87.6 kpsi
x
xy
y
Answer n = S y = 100 = 1.14
σ 87.6
MSS Taking care to note from a quick sketch of Mohr’s circle that one nonzero princi-
pal stress will be positive while the other one will be negative, τ max can be obtained from
the extreme-value shear stress given by Eq. (3–14) without finding the principal stresses.

2
2
σ x − σ y 0 − 40
2
2
τ max = + τ xy = + 45 = 49.2 kpsi
2 2
S y /2 100/2
Answer n = = = 1.02
τ max 49.2
For comparison purposes later in this problem, the nonzero principal stresses can be
obtained from Eq. (3–13) to be 70.0 kpsi and −30 kpsi.
(d) From Eq. (3–13), the nonzero principal stresses are

2
−40 + (−60) −40 − (−60)
2
σ A ,σ B = ± + (15) =−32.0, −68.0 kpsi
2 2
The ordered principal stresses are σ 1 = 0, σ A = σ 2 =−32.0, σ B = σ 3 =−68.0 kpsi.
2
DE σ = (−32) − (−32)(−68) + (−68) 2 1/2 = 59.0 kpsi

Answer n = S y = 100 = 1.70
σ 59.0