Question 5.3: Failure of Cracked Materials Under Static Loading Problem A ...
Failure of Cracked Materials Under Static Loading
Problem A steel support strap designed to hold a 60 000-N static load in axial tension was accidently sawcut during production and now has an edge crack in it. Determine the safety factor of the original, uncracked strap based on yielding and its new “cracked” safety factor based on fracture mechanics. How large could the crack get before it fails? Would heat-treating the part compensate for the loss of strength due to the crack?
Given The material is steel with S_{y} = 540 MPa and K_{c} = 66 MPa-m^{0.5}. The length l = 6 m, width b = 80 mm, and thickness t = 3 mm. The crack width a = 10 mm. The crack is completely through the thickness at one edge of the 80-mm width, similar to Figure 5-19c (p. 269).
Assumptions The load is static and the assembly is at room temperature. The ratio a/b is < 0.13, which allows use of equation 5.14d (p. 269).
K=1.12 \sigma_{\text {nom }} \sqrt{\pi a} \quad a \ll b (5.14d)
1 First calculate the nominal stress in the uncracked part based on total cross section.
\sigma_{n o m}=\frac{P}{A}=\frac{60000}{3(80)}=250 MPa (a)
2 This is a uniaxial stress so is both the principal and the von Mises stress as well. The safety factor against yielding using the distortion energy theory is (Eq 5.8a, p. 251):
N=\frac{S_y}{\sigma^{\prime}} (5.8a)
N_{v m}=\frac{S_y}{\sigma^{\prime}}=\frac{540}{250}=2.16 (b)
3 The stress intensity K at the crack tip can be found for this case from equation 5.14d (p. 269) if the ratio a / b < 0.13:
\quad \frac{a}{b}=\frac{10}{80}=0.125 \\ \text {and} \\ \quad K=1.12 \sigma_{n o m} \sqrt{\pi a}=1.12(250) \sqrt{10 \pi}=49.63 \quad MPa \sqrt{ m } (c)
4 The safety factor against sudden crack propagation is found from equation 5.15 (p. 269).
N_{F M}=\frac{K_c}{K} (5.15)
N_{F M}=\frac{K_c}{K}=\frac{66}{49.63}=1.33 (d)
Note that failure is now predicted to be sudden at a 33% overload, at which point the nominal stress in the part is still below the yield strength. This is too small a safety factor to allow the part to be used in the face of possible sudden fracture.
5 The crack size necessary for failure can be found approximately by substituting K_{c} for K in equation 5.14b (p. 268) and solving for a. The result is a crack of about 18 mm width. Note, however, that the a / b ratio would now exceed that recommended for 10% accuracy with this equation. A more accurate equation for this case could be obtained from one of the references if desired.
K=\beta \sigma_{n o m} \sqrt{\pi a} (5.14b)
6 Assuming the steel has enough carbon to allow heat treatment, through hardening will increase the yield strength but the ductility and the fracture toughness K_{c} will decrease, making the part less safe against a fracture-mechanics failure.
7 The files EX05-03 are on the CD-ROM.