Torsion

144 CHAPTER 5. TORSION
The general solution of the differential equation is g(η) = C1 sinh γη+C2 cosh γη−b2Gκ1, where C1 and C2
are the integration constants that can be evaluated with the help of the boundary conditions, g(η = ±1) = 0.
If follows that g(η) = (cosh γη/ cosh γ − 1) b2Gκ1, and the stress function becomes
cosh γη
Φ(η, ζ) = − 1 (ζ
cosh γ 2 − 1) b 2 Gκ1.
This stress function is shown in fig. 5.17.
Φ
0.25
0.2
0.15
0.1
0.05
0
1
0.5
0
x 3
−0.5
−1
−2
−1
x 2
0
1
2
x 3
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−2 −1.5 −1 −0.5 0
x
2
0.5 1 1.5 2
Figure 5.17: Left figure: stress function Φ. Right figure: distribution of shear stress over cross-section. The
arrows represent the shear stresses; the contours represent constant values of the stress function Φ. a = 2,
b = 1.
For this section bound by a single curve, the externally applied torque is given by eq. (5.73)
M1 = 2 Φ dA = 16ab3
tanh γ
1 − Gκ1.
3 γ
A
Since the sectional torsional stiffness J is defined as the constant of proportionality between the torque and
the twist rate, it follows that
J = 16ab3
tanh γ
1 − G.
3 γ
(5.84)
The stress field is now readily found from the derivatives of the stress function
τ12 = 3 M1
8 ab2 cosh γη/ cosh γ − 1
1 − (tanh γ)/γ
ζ; τ13 = − 3 M1
16 a2b γ sinh γη/ cosh γ
1 − (tanh γ)/γ (ζ2 − 1),
where the twist rate is expressed in terms of the applied torque as κ1 = M1/J. The shear stress distribution
is displayed in fig. 5.17.
Comparison of approximate solutions
The solution of the last two examples are now compared. Instead of using a and b which are the half-lengths
of the cross-section, the full length of the section a ′ = 2a and the width b ′ = 2b will be used instead. The
nondimensional torsional stiffnesses are
J1
a ′ b ′3 =
G 5
18
1
1 + (b ′ /a ′ ;
) 2
J2
a ′ b ′3 =
G 1
1 −
3
tanh γ
. (5.85)
γ
J1 is the torsional stiffness obtained with the crude solution, see eq. (5.83), whereas J2 is that obtained
with the refined solution, see eq. (5.84). For a very thin strip, b → 0 and a/b → ∞. It follows that