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

602
PRESSuRE VESSELS
σ θ
τ max
σ r
45°
σ r = −p i σ θ
σ θ
σ r
p i
σ θ
p o
FIGURE 14.10 Maximum shear stress orientation in a thickwalled
cylinder.
Maximum shear stress: Recall that the maximum shear stress at any point in a solid
material equals one-half of the algebraic difference between the maximum and minimum
principal stresses, which are the circumferential and radial stresses for the thick-walled
cylinder. Thus, at any point in the cylinder, we find that the maximum shear stress is
2 2
1
ab p p
2 ( ) ( i − o)
τmax
= σθ − σ r =
2 2 2
(14.20)
( b − a ) r
The largest value of τ max is found at the inner surface where r = a. The effect of reducing
the outside pressure p o is clearly to increase τ max . Consequently, the largest shear stress
corresponds to r = a and p o = 0. Because σ θ and σ r are principal stresses, τ max occurs on
planes making an angle of 45° with the planes on which σ θ and σ r act. The orientation of
planes of maximum shear stress is shown in Figure 14.10.
Relationships between inner surface and outer surface stresses: If we evaluate the
circumferential stress σ θ on the inside of a thick-walled cylinder at r = a, we find that
2 2
2
b + a
b a p 2b
σ = −
b a p
2 2 2 2
− −
θi i o
Next, we will evaluate σ θ on the outside of the cylinder at r = b to obtain
2
2 2
2a
b a p b + a
σ = −
b a p
2 2 2 2
− −
θo i o
If σ θo is subtracted from σ θi , we discover that the change in circumferential stresses from
the inner surface to the outer surface always equals the difference between the internal and
external pressures:
2 2 2 2
( b − a ) pi
− ( b − a ) po
σθi
− σθ o =
= pi
− p
2 2
o
(14.21)
b − a