Page 258 - Shigley's Mechanical Engineering Design

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Failures Resulting from Static Loading 233
Employing the distortion-energy theory, we find, from Eq. (5–15), that

2 1/2
2
2
2

σ = σ + 3τ zx 1/2 = [(142.6F) + 3(76.4F) ] = 194.5F
x
Equating the von Mises stress to S y , we solve for F and get
Answer F = S y = 81 000 = 416 lbf
194.5 194.5
In this example the strength of the material at point A is S y = 81 kpsi. The strength of
the assembly or component is F = 416 lbf.
Let us apply the MSS theory for comparison. For a point undergoing plane stress
with only one nonzero normal stress and one shear stress, the two nonzero principal
stresses will have opposite signs, and hence the maximum shear stress is obtained from
the Mohr’s circle between them. From Eq. (3–14)

2

2 142.6F
σ x
2
2
τ max = + τ zx = + (76.4F) = 104.5F
2 2
Setting this equal to S y /2, from Eq. (5–3) with n = 1, and solving for F, we get
81 000/2
F = = 388 lbf
104.5
which is about 7 percent less than found for the DE theory. As stated earlier, the MSS
theory is more conservative than the DE theory.

EXAMPLE 5–4 The cantilevered tube shown in Fig. 5–17 is to be made of 2014 aluminum alloy treated
to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size
tube from Table A–8 using a design factor n d = 4. The bending load is F = 1.75 kN,
the axial tension is P = 9.0 kN, and the torsion is T = 72 N · m. What is the realized
factor of safety?

Solution The critical stress element is at point A on the top surface at the wall, where the bend-
ing moment is the largest, and the bending and torsional stresses are at their maximum
values. The critical stress element is shown in Fig. 5–17b. Since the axial stress and
bending stress are both in tension along the x axis, they are additive for the normal
stress, giving

P Mc 9 120(1.75)(d o /2) 9 105d o
σ x = + = + = + (1)
A I A I A I
where, if millimeters are used for the area properties, the stress is in gigapascals.
The torsional stress at the same point is

Tr 72(d o /2) 36d o
τ zx = = = (2)
J J J