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Failures Resulting from Static Loading 247
ability to absorb energy in the plastic region and hence there is a likelihood of brittle
fracture.
The strength-to-stress ratio K Ic /K I can be used as a factor of safety as
K Ic
n = (5–38)
K I
EXAMPLE 5–6 A steel ship deck plate is 30 mm thick and 12 m wide. It is loaded with a nominal uni-
axial tensile stress of 50 MPa. It is operated below its ductile-to-brittle transition tem-
perature with K Ic equal to 28.3 MPa. If a 65-mm-long central transverse crack is
present, estimate the tensile stress at which catastrophic failure will occur. Compare this
stress with the yield strength of 240 MPa for this steel.
Solution For Fig. 5–25, with d = b, 2a = 65 mm and 2b = 12 m, so that d/b = 1 and a/d =
3
65/12(10 ) = 0.00542. Since a/d is so small, β = 1, so that
√ √
−3
K I = σ πa = 50 π(32.5 × 10 ) = 16.0MPa m
From Eq. (5–38),
28.3
K Ic
n = = = 1.77
K I 16.0
The stress at which catastrophic failure occurs is
28.3
K Ic
Answer σ c = σ = (50) = 88.4MPa
K I 16.0
The yield strength is 240 MPa, and catastrophic failure occurs at 88.4/240 = 0.37, or
at 37 percent of yield. The factor of safety in this circumstance is K Ic /K I =
28.3/16 = 1.77 and not 240/50 = 4.8.
EXAMPLE 5–7 A plate of width 1.4 m and length 2.8 m is required to support a tensile force in the
2.8-m direction of 4.0 MN. Inspection procedures will detect only through-thickness
edge cracks larger than 2.7 mm. The two Ti-6AL-4V alloys in Table 5–1 are being con-
sidered for this application, for which the safety factor must be 1.3 and minimum
weight is important. Which alloy should be used?
Solution (a) We elect first to estimate the thickness required to resist yielding. Since σ = P/wt,
we have t = P/wσ. For the weaker alloy, we have, from Table 5–1, S y = 910 MPa.
Thus,
S y 910
σ all = = = 700 MPa
n 1.3
Thus
P 4.0(10) 3
t = = = 4.08 mm or greater
wσ all 1.4(700)