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

Stresses in the outlet Pipe
The longitudinal and circumferential stresses produced in a cylinder by fluid pressure are
given by
pd
pd
σlong
= σhoop
=
4t
2t
where d is the inside diameter of the cylinder and t is the wall thickness. For the outlet pipe,
the wall thickness is t =(6.625 in. − 6.065 in.) 2=0.280 in. The longitudinal stress is
The hoop stress is twice as large:
pd (13psi)(6.065 in.)
σ = = = 70.4 psi
4t 4(0.280 in.)
long Ans.
pd (13psi)(6.065 in.)
σ = = = 140.8 psi
2t 2(0.280 in.)
hoop Ans.
Stress Element at B
The longitudinal axis of the outlet pipe extends in the x direction; therefore, the longitudinal
stress acts in the horizontal direction and the hoop stress acts in the vertical direction
at point B.
Minimum Wall thickness for Standpipe
The maximum hoop stress in the standpipe must be limited to 2,500 psi:
pd
σ hoop = ≤ 2,500 psi
2t
This relationship is then solved for the minimum wall thickness t:
pd (13psi)(108 in.)
t ≥ = = 0.281 in.
2σ
2(2,500 psi)
hoop
Ans.
B
y
140.8 psi
x
70.4 psi
ExAmpLE 14.2
A cylindrical pressure vessel with an outside diameter of
900 mm is constructed by spirally wrapping a 15 mm
thick steel plate and butt-welding the mating edges of the
plate. The butt-welded seams form an angle of 30° with
a transverse plane through the cylinder. Determine the
normal stress σ perpendicular to the weld and the shear
stress τ parallel to the weld when the internal pressure in
the vessel is 2.2 MPa.
y
30°
x
Plan the Solution
After computing the longitudinal and circumferential stresses in the cylinder wall, the
stress transformation equations are used to determine the normal stress perpendicular to
the weld and the shear stress parallel to the weld.
SolutioN
The longitudinal and circumferential stresses produced in a cylinder by fluid pressure are
given by
pd
pd
σlong
= σhoop
=
4t
2t
593