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

be expressed as the product of the differential arclength ds and the tube thickness t; thus,
dA = t ds. Furthermore, the differential arclength can be expressed as ds = r dφ. As a
result, the differential area can be expressed in polar coordinates r and φ as dA = r t dφ.
From these relationships for y and dA, the moment of inertia of the thin-walled tube can
be derived as follows:
2 2 3 2
Iz
= ∫ y dA = ∫ ( r sin φ) rtdφ = rt∫
sin φdφ
3
1 1
= rt
⎡
φ - sinφ cosφ
⎤
⎣⎢ 2 2 ⎦⎥
3
= πrt
2π
0
2π
0
2π
0
z
N.A.
π−θ
r
y dA
dϕ
V
ϕ
θ
ds
y
First Moment of Area, Q
The value of Q can also be determined by integration using polar coordinates.
From the sketch on the left, the value of Q for the area of the cross section above
the arbitrarily chosen sections defined by θ and π - θ will be determined. The
free-body diagram to be considered for the calculation of Q should be symmetric
about the xy plane.
From the definition of Q, the first moment of area dA about the neutral axis
(N.A.) can be expressed as dQ = y dA. Substituting the previous expressions for y
and dA into this definition gives the following expression for dQ in terms of r and φ:
t
dQ = ydA = ( r sin φ)
rt dφ
The angle φ will vary between symmetric limits of θ and π – θ. The following
integration shows the derivation of a general expression for Q:
π-θ
π-θ
2
Q = ∫ dQ = ∫ r t sinφ dφ
θ
2
= rt[ -cosφ]
2
= 2rtcosθ
θ
π-θ
θ
Shear Stress Expressions
The variation of the shear stress τ can now be expressed in terms of the angle θ:
2
VQ V(2r t cos θ)
V
τ = = = cosθ
3
It ( πrt)(2 t)
πrt
Note that the thickness term t in the shear stress equation is the total width of the surface
exposed in cutting the free-body diagram. The free-body diagram considered between
sections at θ and π – θ exposes a total width of two times the wall thickness; hence, the
term 2t appears in the preceding shear stress equation.
For a thin-walled tube in which r .>> t, the cross-sectional area can be approximated
by A ≅ 2π rt. Thus, the shear stress can be expressed as
and the maximum shear stress given by
at a value of θ = 0.
V V
τ =
A/2 cos θ = cosθ
2A
τ max =
2V
A
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