Page 250 - Shigley's Mechanical Engineering Design

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Failures Resulting from Static Loading 225
The distortion-energy theory predicts no failure under hydrostatic stress and agrees
well with all data for ductile behavior. Hence, it is the most widely used theory for duc-
tile materials and is recommended for design problems unless otherwise specified.
One final note concerns the shear yield strength. Consider a case of pure shear τ xy ,
where for plane stress σ x = σ y = 0. For yield, Eq. (5–11) with Eq. (5–15) gives

1/2 S y
2
3τ = S y or τ xy = √ = 0.577S y (5–20)
xy
3
Thus, the shear yield strength predicted by the distortion-energy theory is
(5–21)
S sy = 0.577S y
which as stated earlier, is about 15 percent greater than the 0.5 S y predicted by the MSS
theory. For pure shear, τ xy the principal stresses from Eq. (3–13) are σ A =−σ B = τ xy .
o
The load line for this case is in the third quadrant at an angle of 45 from the σ A ,σ B
axes shown in Fig. 5–9.

EXAMPLE 5–1 A hot-rolled steel has a yield strength of S yt = S yc = 100 kpsi and a true strain at fracture
of ε f = 0.55. Estimate the factor of safety for the following principal stress states:
(a) σ x = 70 kpsi, σ y = 70 kpsi, τ xy = 0 kpsi
(b) σ x = 60 kpsi, σ y = 40 kpsi, τ xy =−15 kpsi
(c) σ x = 0 kpsi, σ y = 40 kpsi, τ xy = 45 kpsi
(d) σ x =−40 kpsi, σ y =−60 kpsi, τ xy = 15 kpsi
(e) σ 1 = 30 kpsi, σ 2 = 30 kpsi, σ 3 = 30 kpsi

Solution Since ε f > 0.05 and S yt and S yc are equal, the material is ductile and both the
distortion-energy (DE) theory and maximum-shear-stress (MSS) theory apply. Both
will be used for comparison. Note that cases a to d are plane stress states.
(a) Since there is no shear stress on this stress element, the normal stresses are
equal to the principal stresses. The ordered principal stresses are σ A = σ 1 = 70,
σ B = σ 2 = 70, σ 3 = 0 kpsi.
DE From Eq. (5–13),

2
2 1/2

σ = [70 − 70(70) + 70 ] = 70 kpsi
From Eq. (5–19),
100
S y
Answer n = = = 1.43
σ 70
MSS Noting that the two nonzero principal stresses are equal, τ max will be from the
largest Mohr’s circle, which will incorporate the third principal stress at zero. From
Eq. (3–16),
70 − 0
σ 1 − σ 3
τ max = = = 35 kpsi
2 2
From Eq. (5–3),
S y /2 100/2
Answer n = = = 1.43
τ max 35