A 12-in-long strip of steel is \frac{1}{8} in thick and 1 in wide, as shown in Fig. 3–28.
If the allowable shear stress is 11 500 psi and the shear modulus is 11.5\left(10^6\right) psi, find the torque corresponding to the allowable shear stress and the angle of twist, in degrees,
(a) using Eq. (3–47) and (b) using Eqs. (3–40) and (3–41).
(a) The length of the median line is 1 in. From Eq. (3–47)
\tau =G\theta _1c = \frac{3T}{Lc^2} ,
T=\frac{L_c^2\tau}{3} =\frac{\left(1\right)\left({1}/{8}\right)^211500 }{3} =59.90\ lbf.in
\theta =\theta _1l=\frac{\tau l}{Gc} =\frac{11500\left(12\right) }{11.5\left(10^6\right)\left({1}/{8}\right) } =0.0960 \ rad=5.5°
A torsional spring rate k_t can be expressed as{T}/{\theta }:\\k_t={59.90}/{0.0960}=624 \ lbf.in/rad
(b) From Eq. (3–40)
\tau _{max} =\frac{T}{abc^2} =\frac{T}{bc^2}\left(3+\frac{1.8}{{b}/{c}} \right)T= \frac{\tau_{max} bc^2 } {3+{1.8}/{ \left({1}/{0.125}\right) } } =55.72 \ lbf.in
From Eq. (3–41)
\theta =\frac{Tl}{\beta bc^3G} \ , \ with \ {b}/{c}={1}/{0.125}=8 ,
\theta =\frac{Tl}{\beta bc^3G} =\frac{55.72\left(12\right) }{0.307\left(1\right)0.125^3\left(11.5\right)10^6 } =0.0970 \ rad=5.6°k_t={55.72}/{0.0970}=574 \ lbf.in/rad
The cross section is not thin, where b should be greater than c by at least a factor
of 10. In estimating the torque, Eq. (3–47) provides a value of 7.5 percent higher than
Eq. (3–40), and is 8.5 percent higher than when the table on page 102 is used.