Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

of the hole has decayed to a value only 7 percent greater than the nominal at a distance of
one diameter from the hole. This rapid decay is typical of the redistribution of stress in the
neighborhood of a discontinuity.
For a ductile material, the stress concentration associated with static loading does not
cause concern, because the material will yield in the region of high stress. With the redistribution
of stress that accompanies this local yielding, equilibrium will be attained and no
fracture will occur. However, if the load is an impact load or a repeated load, instead of a
static load, the material may fracture. Also, if the material is brittle, even a static load may
cause fracture. Therefore, in the case of an impact load or a repeated load on any material,
or static loading on a brittle material, the presence of a stress concentration must not be
ignored.
Specific stress-concentration factors depend on both geometric considerations and the
type of loading. In this section, stress-concentration factors pertaining to axial loading have
been discussed. Stress-concentration factors for torsion and bending will be discussed in
subsequent chapters.
133
STRESS CONCENTRATIONS
ExAmpLE 5.9
The machine part shown is 20 mm thick and is made of C86100
bronze. (See Appendix D for properties.) Determine the maximum
safe load P if a factor of safety of 2.5 with respect to failure by
yield is specified.
SOLUTION
The yield strength of C86100 bronze is 331 MPa. (See Appendix D
for properties.) The allowable stress, based on a factor of safety of
2.5, is 331/2.5 = 132.4 MPa. The maximum stress in the machine
part will occur either in the fillet between the two sections or on the boundary of the
circular hole.
At the Fillet
D
d
From Figure 5.15, K ≅ 1.73. Thus,
= 90 mm
r
60 mm = 1.5 and 15 mm
d
= 60 mm = 0.25
2
s allowAmin
(132.4 N/mm )(60 mm)(20 mm)
P = = = 91,838 N = 91.8 kN
K
1.73
At the Hole
From Figure 5.14, K ≅ 2.36. Thus,
d
D
= 27 mm
90 mm = 0.3
P
27 mm diameter hole
90 mm 60 mm
15 mm radius
P
s allowAnet
P = =
K
Therefore,
2
(132.4 N/mm )(90 mm − 27 mm)(20 mm)
= 70,688 N = 70.7 kN
2.36
Pmax = 70.7 kN
Ans.