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232 Mechanical Engineering Design
For ductile materials with unequal yield strengths, S yt in tension and S yc in com-
pression, the Mohr theory is the best available. However, the theory requires the results
from three separate modes of tests, graphical construction of the failure locus, and fit-
ting the largest Mohr’s circle to the failure locus. The alternative to this is to use the
Coulomb-Mohr theory, which requires only the tensile and compressive yield strengths
and is easily dealt with in equation form.
EXAMPLE 5–3 This example illustrates the use of a failure theory to determine the strength of a mechan-
ical element or component. The example may also clear up any confusion existing
between the phrases strength of a machine part, strength of a material, and strength of
a part at a point.
A certain force F applied at D near the end of the 15-in lever shown in Fig. 5–16,
which is quite similar to a socket wrench, results in certain stresses in the cantilevered
bar OABC. This bar (OABC) is of AISI 1035 steel, forged and heat-treated so that it has
a minimum (ASTM) yield strength of 81 kpsi. We presume that this component would
be of no value after yielding. Thus the force F required to initiate yielding can be
regarded as the strength of the component part. Find this force.
Solution We will assume that lever DC is strong enough and hence not a part of the problem. A 1035
steel, heat-treated, will have a reduction in area of 50 percent or more and hence is a duc-
tile material at normal temperatures. This also means that stress concentration at shoulder
A need not be considered. A stress element at A on the top surface will be subjected to a
tensile bending stress and a torsional stress. This point, on the 1-in-diameter section, is the
weakest section, and governs the strength of the assembly. The two stresses are
M 32M 32(14F)
σ x = = = = 142.6F
3
I/c πd 3 π(1 )
Tr 16T 16(15F)
τ zx = = = = 76.4F
3
J πd 3 π(1 )
Figure 5–16 y
2 in
O
A
12 in
1
1 -in D.
z 2 B
1 -in R. C
8 2 in
1-in D.
15 in
x
F
1
1 -in D.
2
D