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

190
TORSION
Note that the shear flow and the
shear stress always act tangent
to the wall of the tube.
dF = q ds
z
ds
FIGURE 6.22 Deriving a
relationship between internal
torque and shear stress in a
thin-walled section.
y
dA
ρ
O
Median
line
the shear forces V 1 , V 2 , V 3 , and V 4 alone are necessary and sufficient for equilibrium (i.e., no
normal forces are involved). Summing forces in the x direction gives
or
from which it follows that
and, since q = τ × t,
V
= V
1 3
qdx = q dx
1 3
q
τ t
= q
1 3
= τ t
(a)
1 A 3 B
The shear stresses at point A on the longitudinal and transverse planes have the same magnitude;
likewise, the shear stresses at point B on the longitudinal and transverse planes have
the same magnitude. Consequently, Equation (a) may be written as
or
τ t
A A
q
A
= τ t
= q
which demonstrates that the shear flow on a cross section is constant even though the wall
thickness of the section varies. Since q is constant over a cross section, the largest average
shear stress will occur where the wall thickness is the smallest.
Next, an expression relating torque and shear stress will be developed. Consider the
force dF acting through the center of a differential length of perimeter ds, as shown in
Figure 6.22. The differential moment produced by dF about the origin O is simply ρ × dF,
where ρ is the mean radial distance from the perimeter element to the origin. The internal
torque equals the resultant of all of the differential moments; that is,
∫
∫
B B
T = ( dF) ρ = ( qds)
ρ = q ρ ds
This integral may be difficult to integrate by formal calculus; however, the quantity ρ ds is
twice the area of the triangle shown shaded in Figure 6.22, which makes the integral equal
to twice the area A m enclosed by the median line. In other words, A m is the mean area
enclosed within the boundary of the tube wall centerline. The resulting expression relates
the torque T and shear flow q:
B
∫
T = q(2 Am
)
(6.27)
Or, in terms of stress,
T
τ = (6.28)
2At
m
FIGURE 6.23 Thin-walled
shape with an “open” cross
section.
where τ is the average shear stress across the thickness t (and tangent to the perimeter).
The shear stress determined by Equation (6.28) is reasonably accurate when t is relatively
small. For example, in a round tube with a diameter–to–wall-thickness ratio of 20, the
stress as given by Equation (6.28) is 5 percent less than that given by the torsion formula.
It must be emphasized that Equation (6.28) applies only to “closed” sections—that is, sections
with a continuous periphery. If the member were slotted longitudinally (e.g., see
Figure 6.23), the resistance to torsion would be diminished considerably from that of the
closed section.