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(10^{6}) 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. (3–43) and (3–44).
τ = Gθ_{1}c =\frac{3T}{Lc^{2}} (3–47)
τ_{max} =\frac{T}{αbc^{2}} \dot{=} \frac {T}{bc^{2}} (3 +\frac {1.8}{b/c}) (3–43)
where b is the longer side, c the shorter side, and α a factor that is a function of the ratio b/c as shown in the following table.4 The angle of twist is given by
θ =\frac {Tl}{βbc^{3}G} (3–44)
where β is a function of b/c, as shown in the table.
| b/c | 1.00 | 1.50 | 1.75 | 2.00 | 2.50 | 3.00 | 4.00 | 6.00 | 8.00 | 10 | ∞ |
| α | 0.208 | 0.231 | 0.239 | 0.246 | 0.258 | 0.267 | 0.282 | 0.299 | 0.307 | 0.313 | 0.333 |
| β | 0.141 | 0.196 | 0.214 | 0.228 | 0.249 | 0.263 | 0.281 | 0.299 | 0.307 | 0.313 | 0.333 |
