Page 281 - Shigley's Mechanical Engineering Design

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256 Mechanical Engineering Design
Fracture Mechanics
√
p. 243 K I = βσ πa (5–37)

where β is found in Figs. 5–25 to 5–30 (pp. 243 to 246)

K Ic
p. 247 n = (5–38)
K I
where K Ic is found in Table 5–1 (p. 246)

Stochastic Analysis
Mean factor of safety defined as ¯n = μ S /μ σ (μ S and μ σ are mean strength and stress,
respectively)

Normal-Normal Case

2
2
2
2
1 ± 1 − (1 − z C )(1 − z C )
σ
S
p. 250 ¯ n = (5–42)
2
1 − z C 2
S
where z can be found in Table A–10, C S =ˆσ S /μ S , and C σ =ˆσ σ /μ σ .
Lognormal-Lognormal Case

. C n

2
p. 251 ¯ n = exp −z ln(1 + C ) + ln 1 + C 2 = exp C n −z +
n n
2
(5–45)
where
2
C + C 2 σ
S
C n =
1 + C 2
σ
(See other definitions in normal-normal case.)
PROBLEMS
Problems marked with an asterisk (*) are linked to problems in other chapters, as summarized in
Table 1–1 of Sec. 1–16, p. 24.
5–1 A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa.
Using the distortion-energy and maximum-shear-stress theories determine the factors of safety
for the following plane stress states:
(a) σ x = 100 MPa, σ y = 100 MPa
(b) σ x = 100 MPa, σ y = 50 MPa
(c) σ x = 100 MPa, τ xy =−75 MPa
(d) σ x =−50 MPa, σ y =−75 MPa, τ xy =−50 MPa
(e) σ x = 100 MPa, σ y = 20 MPa, τ xy =−20 MPa
5–2 Repeat Prob. 5–1 with the following principal stresses obtained from Eq. (3–13):
(a) σ A = 100 MPa, σ B = 100 MPa
(b) σ A = 100 MPa, σ B =−100 MPa
(c) σ A = 100 MPa, σ B = 50 MPa
(d) σ A = 100 MPa, σ B =−50 MPa
(e) σ A =−50 MPa, σ B =−100 MPa