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28.2 THEORY OF STATIC FAILURE
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For ductile materials, the best estimation method for predicting the onset of yielding, for materials exhibiting equal strengths in tension and compression, is the octahedral shear theory (distortion energy or Hencky-von Mises). The octahedral shear stress is o = 1 3 [( 1 2)2 + ( 2 3)2 + ( 3 1)2]1/2
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STRENGTH UNDER STATIC CIRCUMSTANCES 28.6
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where 1, 2, and 3 are ordered principal stresses (see Chap. 36). In terms of orthogonal stress components in any other directions, the octahedral shear stress is
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2 2 2 o = 1 3 [( x y)2 + ( y z)2 + ( z x)2 + 6( xy + yz + zx)]1/2
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The limiting value of the octahedral shear stress is that which occurs during uniaxial tension at the onset of yield. This limiting value is o = 2Sy 3
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By expressing this in terms of the principal stresses and a design factor, we have 3 1 Sy [( 1 2)2 + ( 2 3)2 + ( 3 1)2]1/2 = = [ o]lim = n 2 2 (28.1)
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The term is called the von Mises stress. It is the uniaxial tensile stress that induces the same octahedral shear (or distortion energy) in the uniaxial tension test specimen as does the triaxial stress state in the actual part. For plane stress, one principal stress is zero. If the larger nonzero principal stress is A and the smaller B, then
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2 2 = ( A + B A B)1/2 =
Sy n
(28.2)
By substituting the relation A,B = we get a more convenient form:
2 = ( x2 + y2 x y + 3 xy)1/2 =
x y 2
x y 2
2 + xy
Sy n
(28.3)
Example 1. A thin-walled pressure cylinder has a tangential stress of and a longitudinal stress of /2. What is the permissible tangential stress for a design factor of n Solution
2 2 = ( A + B A B)1/2
= 2 + From which
Sy n
Sy 3 n
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STRENGTH UNDER STATIC CIRCUMSTANCES 28.7
STRENGTH UNDER STATIC CIRCUMSTANCES
Note especially that this result is larger than the uniaxial yield strength divided by the design factor. Example 2. Estimate the shearing yield strength from the tensile yield strength. Solution. Set A = , B = , and at yield, = Ssy, so
2 = ( A + B2 A B)1/2 2 = [Ssy + ( Ssy)2 Ssy( Ssy)]1/2 = Sy
Solving gives Ssy = Sy = 0.577Sy 3
28.2.1 Brittle Materials To define the criterion of failure for brittle materials as rupture, we require that the fractional reduction in area be less than 0.05; this corresponds to a true strain at fracture of about 0.05. Brittle materials commonly exhibit an ultimate compressive strength significantly larger than their ultimate tensile strength. And unlike with ductile materials, the ultimate torsional strength is approximately equal to the ultimate tensile strength. If A and B are ordered-plane principal stresses, then there are five points on the rupture locus in the A B plane that can be immediately identified (Fig. 28.1). These are
FIGURE 28.1 A B plane with straight-line Coulomb-Mohr strength locus.
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STRENGTH UNDER STATIC CIRCUMSTANCES 28.8
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Locus 1 2: A = Sut, A > B > 0 Point 3: A = Sut = Ssu, B = Sut = Ssu Point 4: A = 0, B = Suc Locus 4 5: B = Suc, A < 0 Connecting points 2, 3, and 4 with straight-line segments defines the modified Mohr theory of failure. This theory evolved from the maximum normal stress theory and the Coulomb-Mohr internal friction theory. We can state this in algebraic terms by defining r = B/ A. The result is A = A = B = Sut n when A > 0, B > A when A > 0, B < 0, r < 1
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