8.1 Several situations in which the possibility of failure is part of the design of a component or
product are as follows: (1) the pull tab on the top of aluminum beverage cans; (2)
aluminum utility/light poles that reside along freeways--a minimum of damage occurs to a
vehicle when it collides with the pole; and (3) in some machinery components, shear pin
are used to connect a gear or pulley to a shaft--the pin is designed shear off before damage
is done to either the shaft or gear in an overload situation.
8.3 This problem asks that we compute the magnitude of the maximum stress that exists at the
tip of an internal crack. Equation (8.1) is employed to solve this problem, as
σm=2σoa
ρt
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
1
/
2
= (2)(170 MPa)
2.5 x10−2mm
2
2.5 x10−4mm
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
1/2
= 2404 MPa (354,000 psi)
8.5 In order to determine whether or not this ceramic material will fail we must compute its
theoretical fracture (or cohesive) strength; if the maximum strength at the tip of the most
severe flaw is greater than this value then fracture will occur--if less than, then there will be
no fracture. The theoretical fracture strength is just E/10 or 25 GPa (3.63 x 106 psi),
inasmuch as E = 250 GPa (36.3 x 106 psi).
The magnitude of the stress at the most severe flaw may be determined using
Equation (8.1) as
σm=2σoa
ρt
= (2)(750 MPa) (0.20 mm)/ 2
0.001mm = 15 GPa 2.2 x 106 psi
(
)
Therefore, fracture will not occur since this value is less than E/10.
8.11W (a) This portion of the problem calls for us to compute the stress at the edge of a circular
through-the-thickness hole in a steel sheet when a tensile stress is applied in a length-wise
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