Page 251 - Mechanical Behavior of Materials

252 Chapter 6 Review of Complex and Principal States of Stress and Strain

|σ 2 − σ 3 | |19.7 − 0|
τ 1 = = = 9.8MPa
2 2

|σ 1 − σ 3 | |100.3 − 0|
τ 2 = = = 50.1MPa
2 2
The maximum shear stress is thus τ 2 , which does not lie in the x-y plane, but acts on planes
◦
inclined 45 to the x-y plane, so that τ max = 50.1MPa (Ans.). This situation was expected,
since the principal stresses σ 1 and σ 2 in the x-y plane are of the same sign.

Example 6.5
A pipe with closed ends has a wall thickness of 10 mm and an inner diameter of 0.60 m. It is
filled with a gas at 20 MPa pressure and is subjected to a torque about its long axis of 1200 kN·m.
Determine the three principal normal stresses and the maximum shear stress. Neglect any effects
of the discontinuity associated with the end closure.
Solution The thin-walled tube approximations of Figs. A.7 and A.8 apply, and the combina-
tion of pressure and torsion give stresses as shown in Fig. E6.5. From thickness t = 10 mm, the
inner, outer, and average radii are, respectively,

r 1 = 300, r 2 = r 1 + t = 310, r avg = r 1 + t/2 = 305 mm

The hoop and longitudinal stresses due to the pressure are then

pr 1 (20 MPa)(300 mm) pr 1
σ t = = = 600 MPa, σ x = = 300 MPa
t 10 mm 2t
σ
t

τ tx σ r

σ
x

Figure E6.5

There is also a radial stress

σ r = 0 (outside), σ r =−p =−20 MPa (inside)