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

(b) Distribution of shear stress
The distribution of shear stress and the direction of the
shear flow is shown in the accompanying figure.
58.5 MPa
78.0 MPa
−78.0 MPa
y
−58.5 MPa
136.5 MPa
z
164.1 MPa
V
43.5 MPa
74.5 MPa
118.0 MPa
−43.5 MPa
−74.5 MPa
ExAmpLE 9.10
A 6061-T6 aluminum thin-walled tube is subjected to a vertical shear force
V = 21,000 lb, as shown in the accompanying figure. The outside diameter of the
tube is D = 8.0 in., and the inside diameter is d = 7.5 in. Plot the distribution of
shear stress in the tube.
y
Plan the Solution
The shear stress distribution in the thin-walled tube will be calculated from the shear
stress formula τ = VQ/It. At the outset, an expression for the moment of inertia of a
thin-walled tube will be derived. From the earlier discussion of shear stresses in
closed thin-walled cross sections, the free-body diagram to be considered for the
calculation of Q should be symmetric about the xy plane. On the basis of this freebody
diagram, the first moment of area, Q, corresponding to an arbitrary location in
the tube wall will be derived and the variation of shear stress will be determined.
z
r
V
t
SolutioN
The shear stress in the tube will be determined from the shear stress formula
τ = VQ/It. The values for both I and Q can be determined by integration using
polar coordinates. Since the tube is thin walled, the radius r of the tube is taken
as the radius to the middle of the tube wall; therefore,
y
dA
ds
r
=
D + d
4
dϕ
ϕ
y
For a thin-walled tube, the radius r is much greater than the wall thickness t
(i.e., r >> t).
z
r
Moment of inertia
From the sketch, observe that the distance y from the z axis to a differential area
dA of the tube wall can be expressed as y = r sin φ. The differential area dA can
t
371