Question 3.5: A tapered bar AB of solid circular cross section is twisted ...
A tapered bar AB of solid circular cross section is twisted by torques T applied at the ends (Fig. 3-19). The diameter of the bar varies linearly from d_{A} at the lefthand end to d_{B} at the right-hand end, with d_{B} assumed to be greater than d_{A}.
(a) Determine the maximum shear stress in the bar.
(b) Derive a formula for the angle of twist of the bar.
(a) Shear stresses. Since the maximum shear stress at any cross section in a solid bar is given by the modified torsion formula (Eq. 3-12), we know immediately that the maximum shear stress occurs at the cross section having the smallest diameter, that is, at end A (see Fig. 3-19):
\tau _{max} =\frac{16T}{\pi d^{3} } (3-12)
\tau _{max} =\frac{16T}{\pi d^{3}_{A} }(b) Angle of twist. Because the torque is constant and the polar moment of inertia varies continuously with the distance x from end A (Case 2), we will use Eq. (3-21) to determine the angle of twist. We begin by setting up an expression for the diameter d at distance x from end A:
\phi =\int_{0}^{L}{d\phi }=\int_{0}^{L}{\frac{Tdx}{GI_{P}\left(x\right) } } (3-21)
d = d_{A} + \frac{d_{B} – d_{A} }{L} x (3-23)
in which L is the length of the bar. We can now write an expression for the polar moment of inertia:
I_{P}\left(x\right) =\frac{\pi d^{4} }{32} =\frac{\pi }{32} \left(d_{A}+\frac{d_{B} – d_{A}}{L} x \right)^{4} (3-24)
Substituting this expression into Eq. (3-21), we get a formula for the angle of twist:
\phi =\int_{0}^{L}{\frac{Tdx}{GI_{P}\left(x\right) } } =\frac{32T}{\pi G} \int_{0}^{L}{\frac{dx}{\left(d_{A}+\frac{d_{B} – d_{A} }{L} x \right)^{4} } } (3-25)
To evaluate the integral in this equation, we note that it is of the form
\int{\frac{dx}{\left(a + bx\right)^{4} } }in which
a = d_{A} b=\frac{d_{B} – d_{A} }{L} (e,f)
With the aid of a table of integrals (see Appendix D available online), we find
\int{\frac{dx}{\left(a + dx\right)^{4} } } =-\frac{1}{3b\left(a + bx\right)^{3} }This integral is evaluated in our case by substituting for x the limits 0 and L and substituting for a and b the expressions in Eqs. (e) and (f). Thus, the integral in Eq. (3-25) equals
\frac{L}{3\left(d_{B} – d_{A} \right) }\left(\frac{1}{d^{3}_{A} }-\frac{1}{d^{3}_{B} } \right) (g)
Replacing the integral in Eq. (3-25) with this expression, we obtain
\phi =\frac{32TL}{3\pi G\left(d_{B} – d_{A} \right) }\left(\frac{1}{d^{3}_{A} } -\frac{1}{d^{3}_{B} } \right) (3-26)
which is the desired equation for the angle of twist of the tapered bar.
Aconvenient form in which to write the preceding equation is
\phi =\frac{TL}{G\left(I_{P} \right)_{A} }\left(\frac{\beta ^{2} +\beta + 1 }{3\beta ^{3} }\right) (3-27)
in which
\beta =\frac{d_{B} }{d_{A} } \left(I_{P} \right)_{A}=\frac{\pi d^{4}_{A} }{32} (3-28)
The quantity β is the ratio of end diameters and \left(I_{P} \right)_{A} is the polar moment of inertia at end A.
In the special case of a prismatic bar, we have β = 1 and Eq. (3-27) gives \phi = TL/G \left(I_{P} \right)_{A}, as expected. For values of β greater than 1, the angle of rotation decreases because the larger diameter at end B produces an increase in the torsional stiffness (as compared to a prismatic bar).