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

31.9 MPa
y
63.8 MPa
t
13.81 MPa
x
30°
39.9 MPa
where d is the inside diameter of the cylinder and t is the wall thickness. The inside diameter
of the cylinder is d = 900 mm − 2(15 mm) = 870 mm. The longitudinal stress in the
tank is
pd (2.2 MPa)(870 mm)
σ long = = = 31.9 MPa
4t 4(15 mm)
The hoop stress is twice as large as the longitudinal stress:
n
pd (2.2 MPa)(870 mm)
σ hoop = = = 63.8 MPa
2t 2(15 mm)
The weld seam is oriented at an angle of 30°, as shown. The normal stress perpendicular
to the weld seam can be determined from Equation (12.3), with θ = −30°:
2 2
σ = σ cos θ + σ sin θ + 2τ sinθcosθ
n x y xy
2 2
= (31.9 MPa)cos (30 − ° ) + (63.8 MPa)sin (30 − ° )
= 39.9 MPa
Ans.
The shear stress parallel to the weld seam can be determined from Equation (12.4):
2 2
τ =−( σ − σ )sinθcos θ + τ (cos θ − sin θ)
nt x y xy
=−(31.9 MPa − 63.8 MPa)sin( − 30 ° )cos( − 30) °
=−13.81 MPa
Ans.
mecmovies
ExAmpLES
m14.3 The pressure tank shown has an outside
diameter of 200 mm and a wall thickness
of 5 mm. The tank has butt-welded seams
forming an angle of β = 25° with a transverse
plane. For an internal gage pressure of
p = 1,500 kPa, determine the normal stress
perpendicular to the weld and the shear stress
parallel to the weld.
m14.4 The strain gage shown
is used to determine the gage
pressure in the cylindrical steel
tank. The tank has an outside
diameter of 1,250 mm and a
wall thickness of 15 mm, and
is made of steel [E = 200 GPa;
ν = 0.32]. The gage is inclined
at a 30° angle with respect to
the longitudinal
axis of the tank.
Determine the
pressure in the
tank corresponding
to a strain gage
reading of 290 µε.
594