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

ExAmpLE 14.3
An open-ended high-pressure steel [E = 200 GPa, ν = 0.3] tube for a boiler has an outside
diameter of 273 mm and a wall thickness of 32 mm. The inside pressure in the boiler tube
is p i = 45 MPa.
(a) Determine the largest radial and circumferential stresses in the tube.
(b) Determine the maximum shear stress in the tube.
(c) Calculate the increase in the inside and outside diameters of the tube after the internal
pressure has been applied.
(d) If the allowable normal stress in the tube material is 140 MPa, determine the
minimum wall thickness required for the boiler tube with an outside diameter of
273 mm.
Plan the Solution
Since the cylinder is subjected to internal pressure only, Equations (14.23) and (14.24)
will be used to compute the normal stresses. From Figure 14.11, we know that the
stress with the largest magnitude will occur on the inner surface of the boiler tube.
The radial stress will be compressive while the circumferential stress will be tensile.
These two stresses are principal stresses. Since the tube is open ended, the longitudinal
stress is zero; thus, the third principal stress is zero and a state of plane stress exists
in the tube. Referring to Figure 14.12, we recognize (a) that the maximum shear
stress can be determined from the two principal stresses and (b) that the planes upon
which the maximum shear stress acts are oriented 45° away from the radial and circumferential
planes.
SolutioN
The outside diameter of the tube is D = 273 mm and the wall thickness is t = 32 mm. The
inside diameter is thus 209 mm. Accordingly, the inner radius is a = 104.5 mm and the
outer radius is b = 136.5 mm.
Radial and Circumferential Stresses
For internal pressure only, the radial and circumferential stresses will be largest on the
inner surface of the tube at r = 104.5 mm. Therefore,
σ =
r
=
2
a pi
b − a
2 2
⎛ b
1 −
⎝
⎜
r
2
(104.5 mm) (45 MPa)
(136.5 mm) − (104.5 mm)
2
2
⎞
⎠
⎟
2 2
⎡ (136.5 mm)
⎢1
−
⎣ (104.5 mm)
2
2
⎤
⎥ 45 MPa
⎦
=− Ans.
σ =
θ
=
2
a pi
b − a
2 2
⎛ b
1 +
⎝
⎜
r
2
(104.5 mm) (45 MPa)
(136.5 mm) − (104.5 mm)
2
2
⎞
⎠
⎟
2 2
⎡ (136.5 mm)
⎢1
+
⎣ (104.5 mm)
2
2
⎤
⎥ 172.44 MPa
⎦
=
Ans.
607