Page 357 - Mechanical Behavior of Materials

358 Chapter 8 Fracture of Cracked Members

Solution (a) Equation 8.15 gives the highest value of K that can be allowed:

√
K Ic 24 MPa m √
K = = = 8.0MPa m
X K 3.0
Hence, this K is employed to obtain the largest crack length that can be permitted:
√ √ √
K = FS g πa, 8.0MPa m = F(100 MPa) πa

Assuming that F = 1 is sufficiently accurate, solving for a gives

a = 2.04 mm Ans.
Since α = a/b = (2.04 mm)/(50 mm) = 0.0408, we are well within the limit for 10% accuracy
on K, and this result is reasonably accurate. If F is allowed to vary, as in Ex. 8.1(c), essentially
the same result is obtained: a = 2.03 mm (Ans.).
(b) In Ex. 8.1(c), the crack length causing failure at the service stress is calculated to be
a c = 16.3 mm. Comparing this with the value of a from part (a) of this example gives the safety
factor on crack length:

X a = a c /a = (16.3mm)/(2.03 mm) = 8.03 Ans.

(c) The safety factor against yielding, calculated as if no crack is present, is given by Eq. 8.18
and is

X o = σ o /S g = (415 MPa)/(100 MPa) = 4.15 Ans.
where the yield strength value is from Table 8.1. A more detailed calculation that uses the fully
plastic limit force as in Eq. 8.19 is

P o 2btσ o (1 − a/b) 2(50 mm)(5mm)(415 MPa) 2.03 mm

X = = = 1 − = 3.98 Ans.
o
P P 50,000 N 50 mm
where the expression for P o is obtained from Fig. A.16(a), and P is the actual service force.
Comments The safety factor on crack length is rather large, as expected. Either of the safety
factors against yielding is higher than X K = 3.0, indicating that this member is closer to brittle
fracture that to yielding; that is, X K = 3.0 is the controlling safety factor.

Example 8.3
An engineering member made of titanium 6Al-4V (annealed) is a plate loaded in tension that
may have a crack in one edge, as shown in Fig. 8.12(c). The applied force is P = 55 kN, the
width is b = 40 mm, and the crack may be as long as a = 6 mm. If a safety factor of 3.0 in stress
is required, what minimum plate thickness t is required?