Page 273 - Shigley's Mechanical Engineering Design

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248 Mechanical Engineering Design
For the stronger alloy, we have, from Table 5–1,
1035
σ all = = 796 MPa
1.3
and so the thickness is
P 4.0(10) 3
Answer t = = = 3.59 mm or greater
wσ all 1.4(796)
(b) Now let us find the thickness required to prevent crack growth. Using Fig. 5–26, we
have
h 2.8/2 a 2.7
= = 1 = = 0.001 93
3
b 1.4 b 1.4(10 )
. √
Corresponding to these ratios we find from Fig. 5–26 that β = 1.1, and K I = 1.1σ πa.
√
K Ic 115 10 3 K Ic
n = = √ , σ = √
K I 1.1σ πa 1.1n πa
√
From Table 5–1, K Ic = 115 MPa m for the weaker of the two alloys. Solving for σ with
n = 1 gives the fracture stress
115
σ = = 1135 MPa
−3
1.1 π(2.7 × 10 )
which is greater than the yield strength of 910 MPa, and so yield strength is the basis
for the geometry decision. For the stronger alloy S y = 1035 MPa, with n = 1 the frac-
ture stress is
K Ic 55
σ = = = 542.9 MPa
−3
nK I 1(1.1) π(2.7 × 10 )
which is less than the yield strength of 1035 MPa. The thickness t is
3
P 4.0(10 )
t = = = 6.84 mm or greater
wσ all 1.4(542.9/1.3)
This example shows that the fracture toughness K Ic limits the geometry when the
stronger alloy is used, and so a thickness of 6.84 mm or larger is required. When the
weaker alloy is used the geometry is limited by the yield strength, giving a thickness of
only 4.08 mm or greater. Thus the weaker alloy leads to a thinner and lighter weight
choice since the failure modes differ.

5–13 Stochastic Analysis 12
Reliability is the probability that machine systems and components will perform their
intended function satisfactorily without failure. Up to this point, discussion in this chap-
ter has been restricted to deterministic relations between static stress, strength, and the
design factor. Stress and strength, however, are statistical in nature and very much tied
to the reliability of the stressed component. Consider the probability density functions

12 Review Chap. 20 before reading this section.